A set of consecutive natural numbers beginning with 1 is written on the board. A student came along and wrote one of these numbers one more time and found the average of the numbers written on the board as 40 20/27.What was the number that was written twice on the board?
27
53
26
60
1)
1.what will be the units digit of (1273)^122! . (in this i knw the concept of cycle and everythin jus have problem understand the factorial part in the power.what is the logic and how do i solve this?)
The answer is
The power 122! is divisible by 4
So it can be written as 4x
=(1237)^4x
3^1=3
3^2=9
3^3=27
3^4=81
So last digit is 1
I am in little bit confusion over this 122! can be a multiple of 5 also.
3^5 remains
and final answer will come as 3...
Can you please explain again?
Find the remainder when 37^(1257) *28^3512 is divided by 5.
2
6
8
4
when 37^(1257) *28^3512 is divided by 5.
the numerator last digit would be 7 * 6 ( how .... look here i approach
37^1257
last digit 7 ^1257( cyclicity is 7,9 3 1)
so last digit would be 7 for 7^1257 similarly for 28 it would be ( 8 4 2 6 )
hence 6
7*6=_2
last digit would be 2
so remainder is 2
PERFECT APPROACH
by option
anwser could be 2/4 not more than 5
for remainder to be 4
the number must have last digit 4 or 9 so eliminate and left is 2
Find the remainder when 37^(1257) *28^3512 is divided by 5.
2
6
8
4
The answer can be found like this ---
1257/4= 1 and 3512/4= 0
so it becomes as 37^1* 28^4 = 7*6 = 42
so the anwer is 2
this is simple as reminder by 5 mean last digit divided by 5.
Thanks
Ashish
The answer can be found like this ---
1257/4= 1 and 3512/4= 0
so it becomes as 37^1* 28^4 = 7*6 = 42
so the anwer is 2
this is simple as reminder by 5 mean last digit divided by 5.
Thanks
Ashish
u sure it would be 42
i dont think so
37*28*28*28*28 =last digit would be 37*96=72
u sure it would be 42
i dont think so
37*28*28*28*28 =last digit would be 37*96=72
hey hey .... I am not talking about last 2 digits... Its just to evaluate last digit. as 7 from first part and 6 is last digit fromm 2nd part .. so last digit of product will be 2.
And last digit by 5 is always reminder in this question.
Hope you get it.
Thanks
Ashish
dil Pe le li tune toh ... chal tc ..and i appreciate ur method thanks
A set of consecutive natural numbers beginning with 1 is written on the board. A student came along and wrote one of these numbers one more time and found the average of the numbers written on the board as 40 20/27.What was the number that was written twice on the board?
27
53
26
60
The answer to this question is 60...
lets understand this way ... we have 5 number begining from 1.
1,2,3,4,5 --- avg is 15/5 =3.
if we take even numbers
1,2,3,4--avg is 2.5
now if the avg is close to .5 then there should be even numbers in row.
and if its odd then its need to be close to .. integer value
As in our case after addtion of an number value is 40.75... so number in original sequence should be 80. which gives 40.5 as avg, and addition of one number makes it 40.75 ...
40.5*80+ NumREP = 40.75* 81
NumREp = 3300.75-3240 = 60.75
so 60 should be answer.
,...........................................................................
full solution..
so the average is 1100/27 ...
let n be total numbers and x be the repeated one...
so (((n*(n+1))/2) +x)/n+1 = 1100/27....so n=26, 53, 80....etc...
solving (without calculating) :
at n=26, x= 1100-(27*13) = very high
at n=53, x=2200-(53*27) = very high i.e., a decreasing function giving reasonable values....
at n=80, x = 60.
therefore x=60
Thanks
Ashish
ashishpatial Saysdil Pe le li tune toh ... chal tc ..and i appreciate ur method thanks
Hey,
Nothing like that yaar...
Its good .
Thanks
Ashish
1) A wooden box (open at the top) of thickness 0.5 cm , length 21 cm , width 11 cm and height 6 cm is painted on the inside . The expenses of painting are Rs 140. What is the rate of painting per square centimetre ?
a) Rs 1.4 b) Rs 0.5 c) Rs 1 d) Rs 2
2) A vertical tower OP stands at the centre O of a square ABCD . Let x and y denote the height of the tower OP and the side of the square AB respectively . Suppose the angle APB = 60 degrees then the relationship between x and y can be expressed as
a) 2y^2 = x^2 b) 2x^2 = y^2 c) 3y^2 = 2x^2 d) 3x^2 = 2y^2
puys please help me solve this ...m not getting the solution ..
Is there any shorter method to solve this one ...
2ndApr2011 SaysIs there any shorter method to solve this one ...
my approach would be as triangle is right angled and ratios are
3 : 4 : 5
so ac 12
cb 16
ab 20
and perpendiculr from c to ab at point m say is 16*12 /20 = 9.6
and 3;4;5
9.6:12.8:16
therefore mb=12.8
now by similarity of triangles cmb and edb
9.6/12.8 =ed/10
ed =7.5 and total area -area of triangle edb = quad
now upper bold part can be avoided as
right angle Triangle formed inside right angle triangle are always in same ratio's that is in this case 3 :4 :5
so u can diractly find ED = 3/4of 10(db )
hence answer would be 58.5
1) A wooden box (open at the top) of thickness 0.5 cm , length 21 cm , width 11 cm and height 6 cm is painted on the inside . The expenses of painting are Rs 140. What is the rate of painting per square centimetre ?
a) Rs 1.4 b) Rs 0.5 c) Rs 1 d) Rs 2
2) A vertical tower OP stands at the centre O of a square ABCD . Let x and y denote the height of the tower OP and the side of the square AB respectively . Suppose the angle APB = 60 degrees then the relationship between x and y can be expressed as
a) 2y^2 = x^2 b) 2x^2 = y^2 c) 3y^2 = 2x^2 d) 3x^2 = 2y^2
puys please help me solve this ...m not getting the solution ..
1)is the ans for the first RS 0.5
2222^7777 + 7777^2222
this question cumes to me ... i would first look at the option
suppose it is 22,99
now 11 is there in both ohk
when u divide 2222 by 11 remainder is zero and if u divide 7777 by 11 it is also zero so
in sum number is divisible by 11
now we need to check whether it is divisible by 2
2222 / 2= remainder 0
7777/2 remainder 1
so we need to find a number that when divided by 2 gives 0 &1 NEVER POSSIBLE
then cum to 9
2222/9 = remainder 8
7777/9= remainder 1
now equation turns into 8^7777 + 1^2222
k
1^2222 divide by 9 always leave 1 remainder
and 8^7777 by cyclicity u can find leave 8 only becoz (cyclicity is 8 &1)
so it will leave 8
and remainder sum would be 8+1 =9
that is divisible by 9 ( forexample if u need to find remainder of 40 by 9 u can write it as 36+4 ohk and 36 is divisible so remainder will be nothing but 0+4 =4)
hence number is divisible by 9
& 99
Hey Ashish
Ur method is fyn...But one thing i want to point out of it is that how come u assume that u shld proceed with checking the answer with 9 despite of not getting a remainder of 0 on both cases :-
2222/9 = remainder 8
7777/9= remainder 1
Because in earlier case of 2 ,u had nt checkd it further on getting 7777/2 remainder 1 ,but in case of 9 u kept solving it.... If its an assumption,then i dnt think dts a wise way to do it in exam..So plz throw sme light on it...
P.S.: Anyone,please correct me if m wrong in the above point. TIA
Hey Ashish
Ur method is fyn...But one thing i want to point out of it is that how come u assume that u shld proceed with checking the answer with 9 despite of not getting a remainder of 0 on both cases :-
2222/9 = remainder 8
7777/9= remainder 1
Because in earlier case of 2 ,u had nt checkd it further on getting 7777/2 remainder 1 ,but in case of 9 u kept solving it.... If its an assumption,then i dnt think dts a wise way to do it in exam..So plz throw sme light on it...
P.S.: Anyone,please correct me if m wrong in the above point. TIA
Hey Sid
Another method would be
2222^7777 + 7777^2222
=(2222^7)^1111 + (7777^2)^1111
The number has to be divisible by 2222^7 + 7777^2( a^n+b^n is divisible by a+b if n is odd)
2222^7 + 7777^2
=1111^2( 2^7*1111^5 + 7^2 )
If we check for 88 and get it right then 44 should also be right, hence 88 cannot be the answer.
We are left with 44, 77 and 99. .
11 divides 1111^2.
So the remaining term should be divisible by either 7,4 or 9
2^3 = -1(mod 9) => 2^7 = 2(mod 9)
1111 = 4(mod 9) => 1111^5 = 4^5(mod 9) = 2^10(mod 9) = -1*2
= -2(mod 9)
2^7*1111^5 = -4(mod 9) and 7^2 = +4( mod 9 )
9 divides ( 2^7*1111^5 + 7^2 )
Hence,99 divides the given number
Coming to your question,( 2^7*1111^5 + 7^2 ) is clearly not divisible by 8, so we are left with 7 and 9.
Now, the first term is clearly not divisible by 7 even though there's a 49.
Any doubts, please revert
Hey Guys
Here r my problems from P&C...Got; screwed up these days in this chapter.. 😞 :- :banghead:
Ques1:- India plays two matches each with NZ & SA.In any match,probability of different outcomes for India is given below:-
OUTCOME WIN LOSS DRAW
Probability 0.5 0.45 0.05
Points 2 0 1
Outcome of all the matches are independent of each other.
Q- What is the probability of India getting at least 7points in contest? Assume SA & NZ play 2 matches.
(a)0.025 (b)0.0875
(c)0.0625 (d)0.975
Q-What is the probability of SA getting at least 4 points? Assume SA & NZ play 2 matches.
(a)0.2025 (b)0.0625
(c)0.0425 (d)Cant be determined
-------------------------------------------------------------------
Ques2:- m men & n women are to be seated in a row so that no two women seats together. If m>n,then find the no. of ways in wich they can b seated.
(a)m!*m+1Pn
(b)m!*m+1Cn
(c)m!*mPn
(d)m!*mCn
-------------------------------------------------------------------
P.S: Please provide answers with Explaination TIA
Hey Sid
Another method would be
2222^7777 + 7777^2222
=(2222^7)^1111 + (7777^2)^1111
The number has to be divisible by 2222^7 + 7777^2( a^n+b^n is divisible by a+b if n is odd)
2222^7 + 7777^2
=1111^2( 2^7*1111^5 + 7^2 )
If we check for 88 and get it right then 44 should also be right, hence 88 cannot be the answer.
We are left with 44, 77 and 99. .
11 divides 1111^2.
So the remaining term should be divisible by either 7,4 or 9
2^3 = -1(mod 9) => 2^7 = 2(mod 9)
1111 = 4(mod 9) => 1111^5 = 4^5(mod 9) = 2^10(mod 9) = -1*2
= -2(mod 9)
2^7*1111^5 = -4(mod 9) and 7^2 = +4( mod 9 )
9 divides ( 2^7*1111^5 + 7^2 )
Hence,99 divides the given number
Coming to your question,( 2^7*1111^5 + 7^2 ) is clearly not divisible by 8, so we are left with 7 and 9.
Now, the first term is clearly not divisible by 7 even though there's a 49.
Any doubts, please revert
The solution u have provided is nice...i got dt...But i was concerning the step occurring in the solution provided by Aashish. In that i want to know hw he suppose to continue solving the problm with case:-9 ..Ur n ashish's solns r diffrnt...
P.S. Correct if m wrong...
Dude because
there are formula's like
a^n+b^n is divisible by a+b if n is odd)
and it doesnt matter how u approach .. the ultimate thing should be ur anwser ...
and also u can deduct any question to any number of levels
here's my explanation:
look there is a theorem called as Chinese remainder theorem
suppose u have to find the remainder of say 435 when divided by 26
u can do it by this way
435/2=1
435/13=6
so u start finder the remainder
13k+6 and 2B+1 form
so the remainder is 19
...
now what i did in that question for 2 is
2222/2 =0
7777/2 = 1
there is no such number possible that will be in two different forms....
moreover y i left there it is because
ODD+even = Always ODD
it could never be Even so it could never be divisible by 2
Q)What will be the remainder when 111111111111111....................729digits is divided by 728?
a)1
b)11
c)111
d)None of these