puys pls help me with these probs

1) The remainder when 10^10+10^100+10^1000+...+10^10000000000 is divided by 7 is
a)0 b)1 c)2 d)5 e)4

2)The remainder when 2^2+22^2+222^2+2222^2+....(222...49 twos)^2 is divisible by 9 is
a)2 b)5 c)6 d)7

3)202x20002x200000002x20000000000000002x2000000...2(31 zeroes) the sum of the digits in this multiplication will be
a)112 b)160 c)144 d)cannot be determined

pls mention the steps as well..thanks!

Dude..i seriously doubt that the system of equations form an ellipse....in fact it forms a square..plz refer the attached fig....

so, the satisfying the equation would be those which lies on the curve formed by all the 4 lines..i.e., square....
now, clearly on one side of the square 4 integral (x,y) pair lies...from the fig..
so for all 4 side..total integral (x,y) pair would be...4*4=16..hence the ans.

if the problem had been x-4 + |y-4 <=4..then the soln would have been all the integral pair (x,y) which lies inside the curve as well as on the line..

do revert in case of any doubts!!

lol... t

x-4|+y-4=4
what does it mean
clearly its the sum of distances of a point (x,y) from lines x=4 and y=4 and the sum is 4

and yeah its should be a square no an ellipse, i just wanted a symmetry! without solving for equations

ah clealry eqn represents a locus of points whose sum of distances from the lines x=4 and y=4 is 4
and that makes it an ellipse centred at 4,4

p.s even if it comes a circle, a circle is also an ellipse

now we take one quadrant and use symmetry for 4

you are looking at individual eqns and i am looking at all the eqns
basically i am looking at the locus

i hope you get it. geometry makes problems quite easier

Dude..i seriously doubt that the system of equations form an ellipse....in fact it forms a square..plz refer the attached fig....

so, the satisfying the equation would be those which lies on the curve formed by all the 4 lines..i.e., square....
now, clearly on one side of the square 4 integral (x,y) pair lies...from the fig..
so for all 4 side..total integral (x,y) pair would be...4*4=16..hence the ans.

if the problem had been x-4 + y-4 <=4..then the soln would have been all the integral pair (x,y) which lies inside the curve as well as on the line..

do revert in case of any doubts!!

implex bhai...dint get u cud u plz expalin a bit in detail....particularly how it can be thought as an ellipse.....cuz according to me its a system of st. lines...

case 1: when x-4 > 0
we can have, x-4 + y-4 = 4 i.e., x+y = 12...when y-4 >0

or, x-4 -y+4 =4 i.e., x-y = 4 ....when y-4 < 0

case 2: when x-4 < 0
we can have, -x+4 + y-4 =4 i.e., -x+y =4..when y-4 >0

or, -x+4 -y+4 =4 i.e., -x-y = -4..when y-4 < 0

ah clealry eqn represents a locus of points whose sum of distances from the lines x=4 and y=4 is 4
and that makes it an ellipse centred at 4,4

p.s even if it comes a circle, a circle is also an ellipse

now we take one quadrant and use symmetry for 4

you are looking at individual eqns and i am looking at all the eqns
basically i am looking at the locus

i hope you get it. geometry makes problems quite easier

havin a lot of probs with number sys!!!!!
can u help????????/

I am getting an ans that is not part of the options...
30^72^82(hope i did not mis-read the problem)
now applyn remainder theorem,
it becomes 8^72^82%11
E(11) = 10
we need to find the remainder of 72^82%10 = R
because 8^(Q.10+R)%11 = 8^Q.10%11 . 8^R%11 = 1.8^R%11
72^82%10 = 2^82%10
div by 2 on dr and nr, 2^81%5 = 2
multiplying by 2, 72^82%10 = 4 (R)
now plugging this back.
8^4%11 = 4
I checked my answer with MS calc. I am getting 4..
Hope this helps

I tried to solve this question using euler's theorem and even i am getting 4

i guess this can be seen as an ellipse
and just check one quadrant and multiply by 4
we get 4*4 =16

implex bhai...dint get u cud u plz expalin a bit in detail....particularly how it can be thought as an ellipse.....cuz according to me its a system of st. lines...

case 1: when x-4 > 0
we can have, x-4 + y-4 = 4 i.e., x+y = 12...when y-4 >0

or, x-4 -y+4 =4 i.e., x-y = 4 ....when y-4 < 0

case 2: when x-4 < 0
we can have, -x+4 + y-4 =4 i.e., -x+y =4..when y-4 >0

or, -x+4 -y+4 =4 i.e., -x-y = -4..when y-4 < 0

one from my side....

if x-4 + y-4 = 4 , then how many integral values can the set (x,y) have?

a)infinite b)5 c)16 d)9 e)25

i guess this can be seen as an ellipse
and just check one quadrant and multiply by 4
we get 4*4 =16

30^72^87 mod 11

72^87/10 >>> 2^87/10 >> 2^3/10 >> 8

so we have...
30^( 10x+8 )/11

we can reduce it to.... 30^8/11 >> -3^8/11 >> 81*81/11 >> 4*4/11

so remainder is 5... as simple as tht...!!!!

(sorry for editing)

flawless approach......ans should be 4 only!!
@vidhyasagar..in top to bottom approach i think u r nor considering the whole no as a single entity.

remainder will definately be 4 if 30^72^82 is divided by 11.
bt hv a luk at the originally posted question....
it says find the remainder wen 30^72^87 is divided by 11.

my ans for this is 5.

explanation-

30^72^87 mod 11 = 8^72^87 mod 11.

now as by euler theorem... 8^10 mod 11= 1

so 8^72^87 can be written as {(8^10)^some number } x 8^8

so[ {(8^10)^some number } x 8^8 ] mod 11=8^8 mod 11

8^8 mod 11 =(-3)^8 mod 11
= (81)^2 mod 11
=4^2 mod 11
=16 mod 11
= 5
hence the remainder is coming out to b 5.

plz correct me if m wrong.

vaibhav

thankyou.

one from my side....

if x-4 + y-4 = 4 , then how many integral values can the set (x,y) have?

a)infinite b)5 c)16 d)9 e)25

the possible combination for first and second terms are (0,4), (1,4)........(4,0)
5 pairs
now each pair has 2 combination on each side which make 4 combination
now for 0 can be only one way so for those two only two ways
4*3+2*2=16

one from my side....

if x-4 + y-4 = 4 , then how many integral values can the set (x,y) have?

a)infinite b)5 c)16 d)9 e)25

I am getting an ans that is not part of the options...
30^72^82(hope i did not mis-read the problem)
now applyn remainder theorem,
it becomes 8^72^82%11
E(11) = 10
we need to find the remainder of 72^82%10 = R
because 8^(Q.10+R)%11 = 8^Q.10%11 . 8^R%11 = 1.8^R%11
72^82%10 = 2^82%10
div by 2 on dr and nr, 2^81%5 = 2
multiplying by 2, 72^82%10 = 4 (R)
now plugging this back.
8^4%11 = 4
I checked my answer with MS calc. I am getting 4..
Hope this helps

flawless approach......ans should be 4 only!!
@vidhyasagar..in top to bottom approach i think u r nor considering the whole no as a single entity.

I don't understand the top to bottom approach..
I hope 30^72^82 is treated as a single number..
if u have 3^7%5, then according to your method it shall be
7%5 = 2
so 3^2%5 = 4
but the answer shall be 2..
Now the place that you are making an error is taking 7%5, or in our problem 72^82%11

now as per Euler, we have (8^10)%11 = 1
so what I am interested is to find the number of powers of 10 that are present in N where (N comes from 8^N)

tht was the reason why I found the remainder with 10..
taking the eg 3^7%5, E(5) = 4
7%4 = 3
3^3%5 = 2

P.S. I am very very confident abt the answer.. even u can try it in calc..

well.... i think i have not given a clear explanation

if u have 3^7%5, then according to your method it shall be
7%5 = 2
so 3^2%5 = 4
but the answer shall be 2..

i dint mean 7%5 has to be done.... if the ques is 3^7^7 then in that case first we have to perform (7^7)/5 which will give 4 as remainder and then proceed as 3^4/5... i hope you get the difference

please go trough page no. 565 of this tread there you will find similar problem and two different approaches as we have... finally vineeth has given a correct method also...

please go tru the same.... and let me know if i got things wrong...

Thanks
Vidhyasagar

Iota bhai....

please elaborate.....

i approached like this

total number of 5 digit numbers that can be formed without repeating the digits will be 5! = 120 numbers

so each of 1,3,5,7,9 will appaer 24 times in all the digits... now i considered only the units digits(i was not able to imagine how it will go for all the places...!!!!!) whose sum will come out as 24*(1+3+5+7+9)...(OK its there in your expression) and (10^5-1)/9 is goin to give me sum for all the places like 10's, 100's,1000's and 10000's.. but i didnt understand how....


i had given my thought process for this... i hope this would help you to clear my doubt better ....

thanks & regards
Vidhyasagar

Each of the positions : units,tens... will be filled by each of the #s 4! times.
so if u consider unit place : it will contain 24 1s,24 3s,24 5s,24 7s & 24 9s. so is with the tens place and so on.
Now hw do u sum up all the digit placed in Units place :
10^0*24*(1+3+5+7+9);
for tens place 10^1*24(1+..+9) and so on
so the sum will be :
24*(25)*(10^4+..+10^0)
and hence the answer.
Hope this clears u r doubt.

as far as i know these kind of problems should be proceeded from top to bottom.....

Check the highlighted part u have done it the other way around...

in this case 30^72^82 first consider 72^82/11 and reduce it into a number of power 1. which can be done as follows....
72^82/11-> 72^2/11 -> 6^2/11 ->3 as remainder

then the expression will become 30^3/11 ->8^3/11 ->6 as the final remainder...

Please check .... somebody please confirm the answer....

Regards
Vidhyasagar

I don't understand the top to bottom approach..
I hope 30^72^82 is treated as a single number..
if u have 3^7%5, then according to your method it shall be
7%5 = 2
so 3^2%5 = 4
but the answer shall be 2..
Now the place that you are making an error is taking 7%5, or in our problem 72^82%11

now as per Euler, we have (8^10)%11 = 1
so what I am interested is to find the number of powers of 10 that are present in N where (N comes from 8^N)

tht was the reason why I found the remainder with 10..
taking the eg 3^7%5, E(5) = 4
7%4 = 3
3^3%5 = 2

P.S. I am very very confident abt the answer.. even u can try it in calc..

I am getting an ans that is not part of the options...
30^72^82(hope i did not mis-read the problem)
now applyn remainder theorem,
it becomes 8^72^82%11
E(11) = 10
we need to find the remainder of 72^82%10 = R
because 8^(Q.10+R)%11 = 8^Q.10%11 . 8^R%11 = 1.8^R%11
72^82%10 = 2^82%10
div by 2 on dr and nr, 2^81%5 = 2
multiplying by 2, 72^82%10 = 4 (R)
now plugging this back.
8^4%11 = 4
I checked my answer with MS calc. I am getting 4..
Hope this helps

as far as i know these kind of problems should be proceeded from top to bottom.....

Check the highlighted part u have done it the other way around...

in this case 30^72^82 first consider 72^82/11 and reduce it into a number of power 1. which can be done as follows....
72^82/11-> 72^2/11 -> 6^2/11 ->3 as remainder

then the expression will become 30^3/11 ->8^3/11 ->6 as the final remainder...

Please check .... somebody please confirm the answer....

Regards
Vidhyasagar

good goin dude
jus saw the ques and also got 7 as the ans
was surprised no1 else even pointed out d mistake to such a easy ques

come on junta
wake up


Originally Posted by chinmay686
) 121212.......upto 300 digits is divided by 99, what is the remainder ?
a)15
b)18
c)12
d)22
Ans:
It has to be div by 9 and 11.
It is div by 9 coz sum of digits is div by 9,
diff btween even and odd digits in the no= 1* 150=150

so 150% 11= 7
so to make it div by 11 we add either 4 or another 11
so the ans is 4 + 11 = 15

HERE THE SOLUTION SHOULD BE 7 AND NOT 15,AS U HAVE TO JUST FIND OUT THE REMAINDER AND NOT THE NUMBER WHICH WILL MAKE IT DIVISIBLE BY THE DIVISOR.SO PLZ CLARIFY UR SOLUTION.


BUT 7 IS NOT IN THE OPTION,FURTHER THE ANSWER IS 7+11=18 AS WE HAVE TO TELL THE REMAINDER WHEN IT IS DIVIDED BY 99 NOT 11.

FOR EG. SAY WE HAVE TO TELL THE REMAINDER WHEN 126 IS DIVIDED BY 15.
CERTAINLY THE ANSWER IS 6 BUT
IF WE SPLIT 15=3x5,THEN WE CAN SEE IT IS DIVISIBLE BY 3 AND THE REMAINDER BY 5 (126/5) WILL BE = 1(ONE)
tHEN WE HAVE TO ADD 5 TO 1 T GET THE RESULT=6

THEREFORE THE SOLUTION TO THE ORIGINAL PROBLEM=7+11=18

Naresh bhai...

i'm getting 6 as the ans. i did the sum using euler no...

on checking i found a silly mistake ....

30^3 wil be 27000 and the final answer would be 6.

please check the same and correct me if i went wrong...

thanks
Vidhyasagar

I am getting an ans that is not part of the options...
30^72^82(hope i did not mis-read the problem)
now applyn remainder theorem,
it becomes 8^72^82%11
E(11) = 10
we need to find the remainder of 72^82%10 = R
because 8^(Q.10+R)%11 = 8^Q.10%11 . 8^R%11 = 1.8^R%11
72^82%10 = 2^82%10
div by 2 on dr and nr, 2^81%5 = 2
multiplying by 2, 72^82%10 = 4 (R)
now plugging this back.
8^4%11 = 4
I checked my answer with MS calc. I am getting 4..
Hope this helps

yeah...vidhya, i edited that silly mistake... Thank u for making me aware..
BTW, how u did...this prob by using Euler's method...plz share.

I am not well versed with the euler theorem....

this is what i have understood by reading different posts... if somebody knows how to use eulertheorem in a better way and save time... then please let us know...

euler no for 11 is 10 {euler number is nothing but the numbers which are smaller and co prime to a given number. here it is 6....in case of a composite number..consider for example 12. here the prime factors will be 2 and 3. and the euler number can be found as follows..... 12*(1-1/2)*(1-1/3) will give u 4 which is the euler number also called as totient}

now comin back t the problem in hand... 30^72^82 / 11

72^82 -> 72^ (8*10 + 2) (72^(8*10))/11 will give 1 as remainder according to euler theorem.

so 72 ^ 82 as as good as 72^2 /11 which will again be reduced into 6^2 /11 and finally 3.

now we have 30^3/11 which clearly gives 6 as remainder on being divided by 11.

hope it helps..

Thanks
Vidhyasagar

yeah...vidhya, i edited that silly mistake... Thank u for making me aware..
BTW, how u did...this prob by using Euler's method...plz share.

Naresh bhai...

i'm getting 6 as the ans. i did the sum using euler no...

on checking i found a silly mistake ....

30^3 wil be 27000 and the final answer would be 6.

please check the same and correct me if i went wrong...

thanks
Vidhyasagar

2) 30^72^82 / 11
first proceed from top exponent..i.e. 72^82 mod 11
same procedure as above like..
72^1 = 72 mod 11 == 6
72^2 = 6*6 mod 11 = 3
72 ^ 4 = 3*3 mod 11 = -2 { i.e. 9-11 =-2}
72^8 = -2*-2 mod 11 = -7
72^16 = -7 * -7 mod 11 = 5
72^32 = 5*5 mod 11 = 3
now for 72^82 mod 11,
we can re-writre it as = {72^32} ^ 2 * 72^16 * 72^2 MOD 11
=> 3*3*5*3 mod 11
=> 135 mod 11
=> 3
30^3 mod 11 = 2700 mod 11
= 5
ans : 5

Naresh bhai...

i'm getting 6 as the ans. i did the sum using euler no...

on checking i found a silly mistake ....

30^3 wil be 27000 and the final answer would be 6.

please check the same and correct me if i went wrong...

thanks
Vidhyasagar

puys pls help me with these probs

1)what is the remainder when 128^1000 is divided by 153?
a)103 b)145 c)118 d)52

i hope this is a relatively shorter method and helps a lot during actual exam....

153=9*17

now 128^1000/9 -> 2^1000/9 ->(2^3)^333 * 2 /9 -> -1*2/9 ->-2 or 7.

similarly

now 128^1000/17 ->2^7000/17 ->(2^4)^1750/17 -> -1.

now check tha answer option which will give remainder 7 on being divided by 9 and remainder 1 on being divided by 17....
clearly that is 52...
hope this method is easy

the # is going to be :
= 4! * (9+7+5+3+1) * (10^5-1)/9
= 6666600

Iota bhai....

please elaborate.....

i approached like this

total number of 5 digit numbers that can be formed without repeating the digits will be 5! = 120 numbers

so each of 1,3,5,7,9 will appaer 24 times in all the digits... now i considered only the units digits(i was not able to imagine how it will go for all the places...!!!!!) whose sum will come out as 24*(1+3+5+7+9)...(OK its there in your expression) and (10^5-1)/9 is goin to give me sum for all the places like 10's, 100's,1000's and 10000's.. but i didnt understand how....


i had given my thought process for this... i hope this would help you to clear my doubt better ....

thanks & regards
Vidhyasagar

1) 128^1000 / 153
proceed this prob as previous 1 which u asked,
128^1 = -25 mod(153) i.e. 128-153
128^2= (-25)^2 = 625 mod 153 = 13
128^4 = 13^2 = 169 mod153 = 16
128^8 = 16^2 = 256 mod 153 = 103
128^16 = 103^2 = 10609 mod 153 = 52
128^32 = 52^2 = 2704 mod 153 = 103
128^64 = 103^2 = 10609 mod 153 = 52
128^128 = 52^2 = 2704 mod 153 = 103
128^256 = 103^2 = 10609 mod 153 = 52
128^512 = 52^2 = 2704 mod 153 = 103
finally for 128^1000 we can re-write it as : 128^512 * 128^256 * 128^128 * 128^64*128^32 * 128^8
which is equals to
= > { 103 * 52 * 103 * 52 * 103 * 103 } mod 153
=> 103 ^2 Mod 153 * 52* 52 * 103 * 103
= > 52 *52*52* 103 ^2 Mod 153
=> { 52 * 52*52* 52 } mod 153
= > 52^2 mod 153 * 52^2
= > 103 * 103 mod 153 ( i.e. similarly for 52^2 mod 153 = 103)
= > 52 is the final ans.

2) 30^72^82 / 11
first proceed from top exponent..i.e. 72^82 mod 11
same procedure as above like..
72^1 = 72 mod 11 == 6
72^2 = 6*6 mod 11 = 3
72 ^ 4 = 3*3 mod 11 = -2 { i.e. 9-11 =-2}
72^8 = -2*-2 mod 11 = -7
72^16 = -7 * -7 mod 11 = 5
72^32 = 5*5 mod 11 = 3
now for 72^82 mod 11,
we can re-writre it as = {72^32} ^ 2 * 72^16 * 72^2 MOD 11
=> 3*3*5*3 mod 11
=> 135 mod 11
=> 3
30^3 mod 11 = 27000 mod 11
= 6
ans : 6

puys pls help me with these probs

1)what is the remainder when 128^1000 is divided by 153?
a)103 b)145 c)118 d)52

2)find the remainder when 30^72^87 is divided by 11
a)5 b)9 c)6 d)3

puys pls help me with these probs



2)find the remainder when 30^72^87 is divided by 11
a)5 b)9 c)6 d)3

T(11)=10
30^72^87%11 come down to
30^(72*87%10)%11=30^4%11=8^4%11=81%11=4
i believe option is wrong!!!!!!!

puys pls help me with these probs

1)what is the remainder when 128^1000 is divided by 153?
a)103 b)145 c)118 d)52

2)find the remainder when 30^72^87 is divided by 11
a)5 b)9 c)6 d)3

T(153)=153*(16/17)(2/3)=3*2*16=96

128^1000%153~=128^40%153 (Since 128^96%153=1)

128%153=-25>>>>>>>>>>>>>>>>>>>>>>> 1
-25^2=625%153=13 >>>>>>>>>>>>>>2
13^2=169%153= 16 >>>>>>>>>>>>>4
16^2= 256%153=-50>>>>>>>>>>>>>> 8
-50^2=2500%153=52 >>>>>>>>>>>16
50*52=-2600%153=1 >>>>>>>>8+16>>>>24
52*1=52 >>>>>>>>>>>>24+16>>>40

may have easier way

The ans is 1 for 3,5,7. For any other triplet a,a+2,a+4 to be prime a should not be divisbl by 3. Hence must leave a rem which will b 2 or 1. Hence a will b of frm 3n+1 or 3n+2. In 1st case a+2 is 3n+3 viz div by 3 hence composite. In 2nd case a+4=3n+6 viz composite. hence a cn nither b multpl of 3 nor cn it leave a rem of 1 nor 2 hence no posbl value 4 a. The only exp a=3

Need some help in solving quite a few probs (cat ones i guess)


8 -> If n is an integer then how many values of n will give an integral value of (16n^2 + 7n + 6)/n
o -> 2,3,4,none

the possible values 1,2,3,6 factors of 6
choice 4

puys pls help me with these probs

1)what is the remainder when 128^1000 is divided by 153?
a)103 b)145 c)118 d)52

2)find the remainder when 30^72^87 is divided by 11
a)5 b)9 c)6 d)3

Need some help in solving quite a few probs (cat ones i guess)


7 -> 5^6 - 1 is divisible by
o -> 13, 31, 5 , none

5^6-1=(5^3-1)(5^3+1)=124*126
124=31*4

answer is 31

Need some help in solving quite a few probs (cat ones i guess)



6 -> It takes the pendulum of a clock 7 sec to strike 4'o clock. How much time will it take to strike 11 o'clock.
o -> 18 sec, 20 s, 19.25 s, 23.33 s

since number of intervals between n bells is n-1
(7/3)*10=70/3=23.33

Need some help in solving quite a few probs (cat ones i guess)


4 -> The product of all integers from 1 to 100 will have the following numbers of zeros at the end
o -> 20,24,19,22

10, 20,..........100 make 11 zeros direct
5,15,25...........when in product make 10 zeros direct as they have 5 in them
now 25, 50 , 75 will make one more zero in the product as they have one more 5 each
11+10+3=24

Need some help in solving quite a few probs (cat ones i guess)



3 -> Suppose one wishes to find distinct positive integers x,y such that (x+y)/xy is also a positive integer. Identify the correct alternative.
o -> Not possible, possible & (x,y) pair is unique, possible and finite no of pairs, possible and infinite pairs

(x+y)

2 -> What is the value of k for which the following system of equations has no solution: 2x-8y=3; and kx+4y=10 ?
o -> -2,1,-1,2

k= -1
because then the equation 2 *-2 will give first equation right side with different

gautam123 Says

Need some help in solving quite a few probs (cat ones i guess)

1 -> If x is positive integer such that 2x+12 is perfectly divisible by x, then the number of possible values of x is?
Options (o) -> 2,5,6,12

for this we must able to take x out that we must be able to right X(2+n)
where n is an integer
not possible values of x are 1,2,3,4,6,12
now answer is 6

2 -> What is the value of k for which the following system of equations has no solution: 2x-8y=3; and kx+4y=10 ?
o -> -2,1,-1,2

3 -> Suppose one wishes to find distinct positive integers x,y such that (x+y)/xy is also a positive integer. Identify the correct alternative.
o -> Not possible, possible & (x,y) pair is unique, possible and finite no of pairs, possible and infinite pairs

4 -> The product of all integers from 1 to 100 will have the following numbers of zeros at the end
o -> 20,24,19,22

5 -> A five digit number is formed using digits 1,3,5,7 and 9 without repeating any one of them. What is the sum of all such possible numbers?
o -> 6666600,6666660,6666666,none

6 -> It takes the pendulum of a clock 7 sec to strike 4'o clock. How much time will it take to strike 11 o'clock.
o -> 18 sec, 20 s, 19.25 s, 23.33 s

7 -> 5^6 - 1 is divisible by
o -> 13, 31, 5 , none

8 -> If n is an integer then how many values of n will give an integral value of (16n^2 + 7n + 6)/n
o -> 2,3,4,none

9 -> ABC is a three digit number in which A>0. The value of ABC is equal to the sum of the factorials of its three digits. What is the value of B?
o -> 9,7,4,2

10 -> If a, a+2, a+4 are prime numbers, then the number of possible solutions for a is ?
o -> 1,2,3,>3

puys pls verify this

i did this problem

find the last 2 digits of 65x29x37x63x71x87x85
a)25 b)35 c)75 d)85

(65*29*37*63*71*87*85) mod 100

Dividing by 5

(13*29*37*63*71*87x17) mod 20

Now, (-7)*9*(-3)*3*(-9)*7*(-3) mod 20

-7^2 * 3^6 *(-3)mod 20 (note -ve sign in front)

Now 7^2 is 49 and we can write 3^6 = 3^3 * 3^3 = 27 * 27

Thus, -9 * 7 * 7*-3 mod 20
Then, -3 * 7 *-3mod 20 (as 9*7 = 63 mod 20 = 3)

-21 mod 20 = -1*-3=3

Thus, to find mod 100, now we multiply by 5...... 3*5 = 15

answer in the book says 85 :huh:..

similar problem is

find the last 2 digits of 65x29x37x63x71x87x62
a)70 b)39 c)10 d)90
in the end we multiply -1*2=-2
-2 mod 20 = 18
Thus, to find mod 100, now we multiply by 5...... 18*5 = 90

answer in the book is 10 :|


pls verify these 2 problems..

oh...damnit, i missed it...such a silly mistake..won't acceptable.:oops:.

avinav2712 Says

Sorry buddy, you got this wrong..... for a = 37, a+2 = 39 which is not prime(divisible by 3) and 57 is also divisible by 3.... Thus, the answer still stays at 1, till any other answers come up

9> ABC is a three digit no in which A>0 . The value of of ABC equal to the sum of the factorials of its three digits. What is the value of B?

Ans--0 fac----0
0 fac ----0
1 fac ----1
2fac ----2
3 fac ----6
4 fac ---- 24
5 fac ---- 120
6 fac ---- 720
7 fac ---- Not a three digit no
For making a three digit no u will need one of N0 6 or 7 or both of them.
But with both No 6 and 7 the no is greater than 800 , and fact 8 is not a 3 digit no
so discard this combi
if u take 6 fac and add something , that also contain no 7 in it and again as fac No 7 not a 3 digit no.
only option is fac 5 + 2 other options..
the ans is 120+ 24+ 1---145

:thumbsup:

Need some help in solving quite a few probs (cat ones i guess)



5 -> A five digit number is formed using digits 1,3,5,7 and 9 without repeating any one of them. What is the sum of all such possible numbers?
o -> 6666600,6666660,6666666,none

the # is going to be :
= 4! * (9+7+5+3+1) * (10^5-1)/9
= 6666600

9 -> ABC is a three digit number in which A>0. The value of ABC is equal to the sum of the factorials of its three digits. What is the value of B?
o -> 9,7,4,2
145 = 1! + 4! + 5!
ANS: 4
10 -> If a, a+2, a+4 are prime numbers, then the number of possible solutions for a is ?
o -> 1,2,3,>3
a, a+2, a+4
Only 3...i.e. for a = 3, 37, 57
ans: 3

Sorry buddy, you got this wrong..... for a = 37, a+2 = 39 which is not prime(divisible by 3) and 57 is also divisible by 3.... Thus, the answer still stays at 1, till any other answers come up

9 -> ABC is a three digit number in which A>0. The value of ABC is equal to the sum of the factorials of its three digits. What is the value of B?
o -> 9,7,4,2
145 = 1! + 4! + 5!
ANS: 4
10 -> If a, a+2, a+4 are prime numbers, then the number of possible solutions for a is ?
o -> 1,2,3,>3
a, a+2, a+4
Only 3...i.e. for a = 3, 37, 57
ans: 3

9 -> ABC is a three digit number in which A>0. The value of ABC is equal to the sum of the factorials of its three digits. What is the value of B?
o -> 9,7,4,2

10 -> If a, a+2, a+4 are prime numbers, then the number of possible solutions for a is ?
o -> 1,2,3,>3

Need some help in solving quite a few probs (cat ones i guess)

1 -> If x is positive integer such that 2x+12 is perfectly divisible by x, then the number of possible values of x is?
Options (o) -> 2,5,6,12

For this, you divide each term of 2x + 12 by x and you are left with 2 + 12/x..... Now, for 2x + 12 to be perfectly divisible by x, 12 has to be divisible by x(since you already have 2 and rest depends on the value of 12/x, which should be an integer)..... Thus, the number of factors of 12 is the answer....1, 2, 3, 4, 6, 12...... Ans is 6

2 -> What is the value of k for which the following system of equations has no solution: 2x-8y=3; and kx+4y=10 ?
o -> -2,1,-1,2

For no solution, the solution is a1/a2 = b1/b2 (not equal to) c1/c2Thus, 2/k = -8/4 but it is not equal to 3/10Thus, from this, k comes out to -1

3 -> Suppose one wishes to find distinct positive integers x,y such that (x+y)/xy is also a positive integer. Identify the correct alternative.
o -> Not possible, possible & (x,y) pair is unique, possible and finite no of pairs, possible and infinite pairs

4 -> The product of all integers from 1 to 100 will have the following numbers of zeros at the end
o -> 20,24,19,22

5 -> A five digit number is formed using digits 1,3,5,7 and 9 without repeating any one of them. What is the sum of all such possible numbers?
o -> 6666600,6666660,6666666,none

6 -> It takes the pendulum of a clock 7 sec to strike 4'o clock. How much time will it take to strike 11 o'clock.
o -> 18 sec, 20 s, 19.25 s, 23.33 s

At 4 o' clock, the clock would strike 4 times, with the first strike at 4 itself and the next 3 strikes later..... Thus, time taken per strike.... 7/3(and not 7/4)......Thus, at 11 o' clock, there'll be 10 strikes after the initial one and thus the total time taken would be...... 7/3 * 10 = 70/3 = 23.33 s

7 -> 5^6 - 1 is divisible by
o -> 13, 31, 5 , none

For this, you've to ensure that 5^6 has a remainder of 1 when divided by either 13, 31 etc..... Now checking for 135^2 = -1 mod 13 (5^2 = 25 and 26 is a multiple of 13, thus a remainder of 12 or -1, taking it as -1 makes things easier)(5^2)^3 = 5^6 = (-1)^3 mod 13 = -1 mod 13 = 12 mod 13, which we don't wantNext, moving to 315^2 = -6 mod 31(5^2)^2 = 5^4 = (-6)^2 mod 31 = 5 mod 31(as -6^2 = 36)5^2 * 5^4 = 5^6 = 5 * (-6) mod 31 = -30 mod 31 = 1 mod 31....(multiplying results of 5^2 and 5^4)thus, 31 is the answer

8 -> If n is an integer then how many values of n will give an integral value of (16n^2 + 7n + 6)/n
o -> 2,3,4,none

Similar logic for this as well as with the 1st ques..... finally you have to find the no. of factors of 6.... 1, 2, 3, 6...... 4 is the answer

9 -> ABC is a three digit number in which A>0. The value of ABC is equal to the sum of the factorials of its three digits. What is the value of B?
o -> 9,7,4,2

10 -> If a, a+2, a+4 are prime numbers, then the number of possible solutions for a is ?
o -> 1,2,3,>3

I'm not very sure of this last one..... but I think the answer is 1.....it being (3, 5, 7).... I don't think you'd find any other example like this, but I may be wrong on this..... didn't think enough on this one

Need some help in solving quite a few probs (cat ones i guess)

1 -> If x is positive integer such that 2x+12 is perfectly divisible by x, then the number of possible values of x is?
Options (o) -> 2,5,6,12

2 -> What is the value of k for which the following system of equations has no solution: 2x-8y=3; and kx+4y=10 ?
o -> -2,1,-1,2

3 -> Suppose one wishes to find distinct positive integers x,y such that (x+y)/xy is also a positive integer. Identify the correct alternative.
o -> Not possible, possible & (x,y) pair is unique, possible and finite no of pairs, possible and infinite pairs

4 -> The product of all integers from 1 to 100 will have the following numbers of zeros at the end
o -> 20,24,19,22

5 -> A five digit number is formed using digits 1,3,5,7 and 9 without repeating any one of them. What is the sum of all such possible numbers?
o -> 6666600,6666660,6666666,none

6 -> It takes the pendulum of a clock 7 sec to strike 4'o clock. How much time will it take to strike 11 o'clock.
o -> 18 sec, 20 s, 19.25 s, 23.33 s

7 -> 5^6 - 1 is divisible by
o -> 13, 31, 5 , none

8 -> If n is an integer then how many values of n will give an integral value of (16n^2 + 7n + 6)/n
o -> 2,3,4,none

9 -> ABC is a three digit number in which A>0. The value of ABC is equal to the sum of the factorials of its three digits. What is the value of B?
o -> 9,7,4,2

10 -> If a, a+2, a+4 are prime numbers, then the number of possible solutions for a is ?
o -> 1,2,3,>3

trevis Says

thanks a lot bro !!

Just the "Thanks" button would have done.... No need for a separate post buddy.....

thanks a lot bro !!

bx-ay = bk ---- 1
cy-bz = ck ---- 2
az-cx = ak ---- 3

from eq.3 we get x,
as, a(z-k)/c = x

from eq.1,

b( a(z-k)/c) - a( bz+ck)/c = bk
ab(z-k)- a(bz+ck) = bck
abz-abk-abz-ack = bck
-abk-ack = bck
cancel k both sides...
bc = -(ac+ab)
i.e. ab+bc+ca= 0
so ans: C

if (bx-ay)/b =(cy-bz)/c = (az-cx)/a and bx-ay NE 0 and cy-bz NE 0 and az-cx NE 0 ..then whcih of the following is true .??
a. a+b+c = 0
b. a+b+-c = 0
c. ab+bc+ca = 0
d. ab+bc-ca= 0
ps: NE is Not Equal to

if (bx-ay)/b =(cy-bz)/c = (az-cx)/a and bx-ay NE 0 and cy-bz NE 0 and az-cx NE 0 ..then whcih of the following is true .??
a. a+b+c = 0
b. a+b+-c = 0
c. ab+bc+ca = 0
d. ab+bc-ca= 0
ps: NE is Not Equal to

yea, accepted that is not the formulae for the Sum of divisors but its for the unique divisors ...OMG! what a kind of mess...:)..
Thanks Vaibhav & Sandeep....for makin me aware of my mistake..:)
Finally, plz find the formulae i'm attached below for the sum of the divisors:
i hope you might find clear explanation now:
First check one Eg:
0(12) is the number of the divisors of 12:
0(12) = 10 + 20 + 30 + 40 + 60 + 120
= 1 + 1 + 1 + 1 + 1 + 1 = 6.
while 1(12) is the sum of all the divisors:
1(12) = 1^1 + 2^1 + 3^1 + 4^1 + 6^1 + 12^1
= 1 + 2 + 3 + 4 + 6 + 12 = 28.
sum of its divisors equals :
2^2 * 3 = {(2^3 - 1)/2-1 } * { (3^2 -1)/3-1} = 7*4 = 28

now similary, following the above principle for a= 2^66*3^33 , b= 2^23*17^23, c= 2^47
to find the total sum of divisors of a + b + c,
1st we calculate individually a, b, c and add them..
for a: 2^66*3^33
= {(2^67 - 1)/2-1 } * { (3^34 -1)/3-1} = {2^67 - 1}*{3^34-1} /2
for b: 2^23*17^23,
= {(2^24 - 1)/2-1 } * { (17^24 -1)/17-1} = {(2^24 - 1) } * { (17^24 -1)/16}
for c: 2^47
= {(2^48 - 1)/2-1 } = {(2^48 - 1)}
finally,
total sum of its divisors of a + b + c =
i.e. {2^67 - 1}*{3^34-1} /2 + {2^24 - 1} * { (17^24 -1)/16} + {2^48 - 1}
=> {{2^70 - 8}*{3^34 - 1} + {2^24 - 1} * { (17^24 -1)} + {2^52 - 16} } /16

well, i hope the Method is correct now...

sir ji everything is clear now except onething....that is to find the total sum of of divisors of expression a+b+c..y do v add the individual sum of each of a, b and c...

yea, accepted that is not the formulae for the Sum of divisors but its for the unique divisors ...OMG! what a kind of mess...:)..
Thanks Vaibhav & Sandeep....for makin me aware of my mistake..:)
Finally, plz find the formulae i'm attached below for the sum of the divisors:
i hope you might find clear explanation now:
First check one Eg:
0(12) is the number of the divisors of 12:
0(12) = 10 + 20 + 30 + 40 + 60 + 120
= 1 + 1 + 1 + 1 + 1 + 1 = 6.
while 1(12) is the sum of all the divisors:
1(12) = 1^1 + 2^1 + 3^1 + 4^1 + 6^1 + 12^1
= 1 + 2 + 3 + 4 + 6 + 12 = 28.
sum of its divisors equals :
2^2 * 3 = {(2^3 - 1)/2-1 } * { (3^2 -1)/3-1} = 7*4 = 28

now similary, following the above principle for a= 2^66*3^33 , b= 2^23*17^23, c= 2^47
to find the total sum of divisors of a + b + c,
1st we calculate individually a, b, c and add them..
for a: 2^66*3^33
= {(2^67 - 1)/2-1 } * { (3^34 -1)/3-1} = {2^67 - 1}*{3^34-1} /2
for b: 2^23*17^23,
= {(2^24 - 1)/2-1 } * { (17^24 -1)/17-1} = {(2^24 - 1) } * { (17^24 -1)/16}
for c: 2^47
= {(2^48 - 1)/2-1 } = {(2^48 - 1)}
finally,
total sum of its divisors of a + b + c =
i.e. {2^67 - 1}*{3^34-1} /2 + {2^24 - 1} * { (17^24 -1)/16} + {2^48 - 1}
=> {{2^70 - 8}*{3^34 - 1} + {2^24 - 1} * { (17^24 -1)} + {2^52 - 16} } /16

well, i hope the Method is correct now...

exactly.....this is the confusion causing part....
also...
let a= 2^66*3^33 , b= 2^23*17^23, c= 2^47


can u plz explain y the sum of divisors of a, b and c will give u the total number of divisors for the entire term....

@vaibhav,

the total number of divisors are
n(A)+n(B)+n(C)+ n(AXB)+ n(AxC)+ n(BXC)+ n(AxBxC)
TOTAL SUM OF DIVISORS...= 6 + 5+ 2 + 30+12+10+60 = 125

NOW USING ANOTHER METHOD:
I.E. =Total no: of divisors in an expansion a^p * b^q * c^r is (p+1)(q+1)(r+1) - 1.
so, we can apply to this..
i.e. = (6+1)*(5+1)*(2+1) - 1 = 126-1 = 125
matching..right!

we can check for our problem too..i.e. 2^66*3^33 + 2^23*17^23 + 2^47
{67*34 - 1 }+ {24*24 - 1} + {48 - 1}
= 2277+575+47 = 2899

Hope now its clear for U! :)

hey plz check the bold part ...u r writing no. of divisors and finding sum of divisors and also in the second bold part the total no of divisors formula seems incorrect to me....(why it shld be (p+1)..(r+1)-1)
plz explain how u got the formula!!

exactly.....this is the confusion causing part....
also...
let a= 2^66*3^33 , b= 2^23*17^23, c= 2^47


can u plz explain y the sum of divisors of a, b and c will give u the total number of divisors for the entire term....

@vaibhav,

the total number of divisors are
n(A)+n(B)+n(C)+ n(AXB)+ n(AxC)+ n(BXC)+ n(AxBxC)
TOTAL SUM OF DIVISORS...= 6 + 5+ 2 + 30+12+10+60 = 125

NOW USING ANOTHER METHOD:
I.E. =Total no: of divisors in an expansion a^p * b^q * c^r is (p+1)(q+1)(r+1) - 1.
so, we can apply to this..
i.e. = (6+1)*(5+1)*(2+1) - 1 = 126-1 = 125
matching..right!

we can check for our problem too..i.e. 2^66*3^33 + 2^23*17^23 + 2^47
{67*34 - 1 }+ {24*24 - 1} + {48 - 1}
= 2277+575+47 = 2899

Hope now its clear for U! :)

hey plz check the bold part ...u r writing no. of divisors and finding sum of divisors and also in the second bold part the total no of divisors formula seems incorrect to me....(why it shld be (p+1)..(r+1)-1)
plz explain how u got the formula!!

@vaibhav,
i proved the result with 2 methods...plz chek below
eg: as 2^6 3^5 5^2
A={2,4,8..} n(A)=6
B={3,9,27,.} n(B)=5
C={5,25} n(C)=2
the total number of divisors are
n(A)+n(B)+n(C)+ n(AXB)+ n(AxC)+ n(BXC)+ n(AxBxC)
TOTAL SUM OF DIVISORS...= 6 + 5+ 2 + 30+12+10+60 = 125

NOW USING ANOTHER METHOD:
I.E. =Total no: of divisors in an expansion a^p * b^q * c^r is (p+1)(q+1)(r+1) - 1.
so, we can apply to this..
i.e. = (6+1)*(5+1)*(2+1) - 1 = 126-1 = 125
matching..right!

we can check for our problem too..i.e. 2^66*3^33 + 2^23*17^23 + 2^47
{67*34 - 1 }+ {24*24 - 1} + {48 - 1}
= 2277+575+47 = 2899

Hope now its clear for U! :)

take eg: as 2^6 3^5 5^2
A={2,4,8..} n(A)=6
B={3,9,27,.} n(B)=5
C={5,25} n(C)=2
the total number of divisors are
n(A)+n(B)+n(C)+ n(AXB)+ n(AxC)+ n(BXC)+ n(AxBxC)
where AXB are number formed by multipliying each element of A for each element of B, and taking into account
that aixbi = bixai
for instance n(AXB)=30
apply the same funda here...
so finally total no: of divisiors for:
2^66*3^33 + 2^23*17^23 + 2^47
|
= {66+33+ 66*33} + {23+23+23*23} + 47
= 2178+99 + 46+ 529+47 = 2899

i hope u got now...?

but there is another formula for calculating the number of divisors...
i.e if N= a^p x b^q....
then the total number of divisors are...(p+1)(q+1)

for ex-N= 2^6 X 3^5..
THEN no of divisors is ....(6+1)(5+1)=42

but going by ur method...the no of divisors cm out to 6+5+6x5=41...
also here...
2^66*3^33 + 2^23*17^23 + 2^47
|
= {66+33+ 66*33} + {23+23+23*23} + 47
= 2178+99 + 46+ 529+47 = 2899

can u explain the funda of getting the total number of divisors by adding total number of divisors for each term...

take eg: as 2^6 3^5 5^2
A={2,4,8..} n(A)=6
B={3,9,27,.} n(B)=5
C={5,25} n(C)=2
the total number of divisors are
n(A)+n(B)+n(C)+ n(AXB)+ n(AxC)+ n(BXC)+ n(AxBxC)
where AXB are number formed by multipliying each element of A for each element of B, and taking into account
that aixbi = bixai
for instance n(AXB)=30
apply the same funda here...
so finally total no: of divisiors for:
2^66*3^33 + 2^23*17^23 + 2^47
|
= {66+33+ 66*33} + {23+23+23*23} + 47
= 2178+99 + 46+ 529+47 = 2899

i hope u got now...?

but there is another formula for calculating the number of divisors...
i.e if N= a^p x b^q....
then the total number of divisors are...(p+1)(q+1)

for ex-N= 2^6 X 3^5..
THEN no of divisors is ....(6+1)(5+1)=42

but going by ur method...the no of divisors cm out to 6+5+6x5=41...
also here...
2^66*3^33 + 2^23*17^23 + 2^47
|
= {66+33+ 66*33} + {23+23+23*23} + 47
= 2178+99 + 46+ 529+47 = 2899

can u explain the funda of getting the total number of divisors by adding total number of divisors for each term...

can someone explain this more clearly?
iunderstand till this 32^32^32%7~=4^32^32
after this pls explain..
how is 4^32^32 = 2^16^32 ??

The other way of doing this question can be,
1) divide 32 by 7, rem is 4 so we have to find the rem when 4^32^32 div by 7.
2) Now find the power of 4 which when divided by 7 will give the rem of 1, 4^3 here
3) now divide 32^32 with this power i.e. 3 and find the rem, it is 1
4) hence now the prob reduces to finding the rem of 4^3k+1
5) since 4^3k will give a rem of 1 and 4 by 7 will give a rem of 4.

hence final ans is 4.

@ getintoiimb,

4)remainder when (32^32^32)/7
= {32}^32^32 =
start from top exponent....

32^2 mod 7 = 4^2 mod 7 = 16 mod 7 = 2
32^4 mod 7 = 2^2 mod 7 = 4 mod 7 = 4
32^8 mod 7 = 4^2 mod 7 = 16 mod 7 = 2
32^16 mod 7 = 2^2 mod 7 = 4 mod 7 = 4
32^32 mod 7 = 4^2 mod 7 = 16 mod 7 = 2

so , 2
(32^32)^2 mod 7 = 2^2 mod 7 = 4 mod 7 = 4
(32^32)^4 mod 7 = 4^2 mod 7 =16 mod 7 = 2
(32^32)^8 mod 7 = 2^2 mod 7 = 4 mod 7 = 4
(32^32)^16 mod 7 = 4^2 mod 7 =16 mod 7 = 2
(32^32)^32 mod 7 = 2^2 mod 7 = 4 mod 7 = 4

so ans is : 4

can someone explain this more clearly?
iunderstand till this 32^32^32%7~=4^32^32
after this pls explain..
how is 4^32^32 = 2^16^32 ??

take eg: as 2^6 3^5 5^2
A={2,4,8..} n(A)=6
B={3,9,27,.} n(B)=5
C={5,25} n(C)=2
the total number of divisors are
n(A)+n(B)+n(C)+ n(AXB)+ n(AxC)+ n(BXC)+ n(AxBxC)
where AXB are number formed by multipliying each element of A for each element of B, and taking into account
that aixbi = bixai
for instance n(AXB)=30
apply the same funda here...
so finally total no: of divisiors for:
2^66*3^33 + 2^23*17^23 + 2^47

= {66+33+ 66*33} + {23+23+23*23} + 47
= 2178+99 + 46+ 529+47 = 2899

i hope u got now...?

1) sum of the divisors of the above also represented as:
= 2^66*3^33 + 2^23*17^23 + 2^47
= {66+33+ 66*33} + {23+23+23*23} + 47

can u explain the above step plz.......

32%7=4
4^2%7=2
32^32^32%7~=4^32^32
~=2^16^32
~=4^8^32
~2^4^32
~4^2^32
~2^32
~2^8
~2^2=4

can someone explain this more clearly?
iunderstand till this 32^32^32%7~=4^32^32
after this pls explain..
how is 4^32^32 = 2^16^32 ??

1) sum of the divisors of the above also represented as:
= 2^66*3^33 + 2^23*17^23 + 2^47
= {66+33+ 66*33} + {23+23+23*23} + 47

can u explain the above step plz.......

1) sum of the divisors of the above also represented as:
= 2^66*3^33 + 2^23*17^23 + 2^47
= {66+33+ 66*33} + {23+23+23*23} + 47
= 2178+99 + 46+ 529+47 = 2899

2) last 2 digits of the expression 65*29*37*63*71*87*62
its..90 easy
3)Remainder when 9 + 9^3 + 9^5 +.....9^2n+1 is divided by 6
9^1 / 6 = 3
9^2 / 6 = 3
9^3 / 6 = 3
9^4 / 6 = 3
9^5 / 6 = 3
..so on..
..finally we have: 3+ 3+3+..........................2n+1 times..
3*{Sigma of (1+1+......2n+1)....} = 3*( 2n+1)/6 = n+1/2 is the remainder....
well, options could have made little help....let me know actual ans.

4)Remainder when (1!)^3 + (2!)^3 +......+(1152!)^3 is divided by 1152
Ans: 225
1!^3 = 1
2!^3 = 8
3!^3 = 216
except these above three...i.e.4!^3 = 13824, 5!^3.......all are divisible by 1152...so, required remainders are are: 1+8+216 = 225

ANY please solve the following qsns...!

1) total no of divisors of 12^33 + 34^23 + 2^47
sum of the divisors of the above also

2) last 2 digits of the expression 65*29*37*63*71*87*62

3)Remainder when 9 + 9^3 + 9^3 +.....9^2n+1 is divided by 6

4)Remainder when (1!)^3 + (2!)^3 +......+(1152!)^3 is divided by 1152