Find the units digit of (1)^5^5 + (2)^5^5 + (3)^5^5 +.....+ (99)^5^5 ?

a)1 b) 0 c)2 d) 4

is it 0????
1 wl have last digit 1
2 lw hv to ... so on 9 wl hav 9
so sum wl b ( 1+2+3+...+9+0) *10 which wl have last digit 0

:biggrin:

Find the units digit of (1)^5^5 + (2)^5^5 + (3)^5^5 +.....+ (99)^5^5 ?

a)1 b) 0 c)2 d) 4

0

1^5^5 + 9^5^5=0
2^5^5+9^5^5=0
same for 3,7 and 4,6
now the number that ends with 5 will always give5 as its last digit
there 10 such numbers so 5*10=0
unit digit is 0

jain4444 Says

how many positive two-digits integers are factors of (2^24 - 1) ?

wrong post.........

jain4444 Says

How many positive two-digits integers are factors of (2^24 - 1) ?

Just a though process, not solution

2^24 -1 => odd so no positive even integer should be considered.

divisible by 13 as E(13) =12 (1-1 =0)
should not be divisible by 26 although as it is even.

17 => E(17) =16

2^8-1 => 4*4*4*4 - 1 => -1*-1 -1 (should be divisible by 17)

But it can be written as
(2^12+1)(2^12-1)
(2^12+1)(2^6+1)(2^6-1)

(2^12+1)(2^6+1)(2^3+1) (2^3-1)

= 4097*65*9*7
= 17*241*13*5*3^2*7

= so 2*2*2*2*3*2 = 96 factors out of these 1,3,5,7,9,241...have to be removed or

13,17,85,65,54,15,63,39,51,45 = so total 10

EDIT: 91, 35

jain4444 Says

How many positive two-digits integers are factors of (2^24 - 1) ?

(2^24-1) = (2^12+1)(2^6+1)(2^3+1)(2^3-1)
=> 4097*65*9*7
=>17*241*5*13*9*7
=>3^2*5*7*13*17*241

2 digit factors => 13, 17, 15, 45, 21, 63, 39, 51, 35, 65, 85, 91, =>12 factors

Let g(x) be a function such that g(x+1)+g(x-1)=g(x) for every real x. Then for what value of p is the relation g(x+p)=g(x) necessarily true for every real x?

1) 5
2) 3
3) 2
4) 6
5) none of these

g(x) = g(x+1) + g(x-1)
g(x+1) = g(x) + g(x+2)
adding together we get
g(x) + g(x+1) = g(x) + g(x+1) + g(x-1) + g(x+2)
g(x+2) = -g(x-1)
g(x) = -g(x+3)
g(x+3) = -g(x+6)
so g(x) = g(x+6)

Hence value for p=6

jain4444 Says

How many positive two-digits integers are factors of (2^24 - 1) ?

is it 12.....
2^24 -1
(2^12 + 1)(2^6 +1)(2^3 + 1)(2^3 -1)
4097 x 65 x 9 x 7
3^2 x 5 x 7 x 13 x 17 x 241
so 2 digit factors can be
13.17,15,21,39,51,45,63,35,65,85,91...
point out if i have missed some...

Find the units digit of (1)^5^5 + (2)^5^5 + (3)^5^5 +.....+ (99)^5^5 ?

a)1 b) 0 c)2 d) 4

0

jain4444 Says

How many positive two-digits integers are factors of (2^24 - 1) ?

I am not sure but is the answer 9????

Let g(x) be a function such that g(x+1)+g(x-1)=g(x) for every real x. Then for what value of p is the relation g(x+p)=g(x) necessarily true for every real x?
1) 5
2) 3
3) 2
4) 6
5) none of these

Should be 6.
f(2) = f(1)+f(3)

means middle no is sum of baju baju wale no

1 3 2 -1 -3 -2 1 3 2 -1...........
so after every 6 values, chain is repeated so my take 6.

Find the units digit of (1)^5^5 + (2)^5^5 + (3)^5^5 +.....+ (99)^5^5 ?

a)1 b) 0 c)2 d) 4

is it 0????

How many positive two-digits integers are factors of (2^24 - 1) ?

Let g(x) be a function such that g(x+1)+g(x-1)=g(x) for every real x. Then for what value of p is the relation g(x+p)=g(x) necessarily true for every real x?

1) 5
2) 3
3) 2
4) 6
5) none of these

g(1) + g(-1) = g(0)

g(2) + g(0) = g(1)

g(2) = - g(-1)

so p =3

g(2)+g(0) = g(1)
g(3)+g(1) = g(2)
=>g(1) = g(3)-g(2)

=>g(3)-g(2) = g(2)+g(0)
=> g(3)=g(0)
=> g(0+3) = g(0)

so p = 3

how come g(1)=g(3)-g(2)
it should be g(2)-g(3) rite???

Find the units digit of (1)^5^5 + (2)^5^5 + (3)^5^5 +.....+ (99)^5^5 ?

a)1 b) 0 c)2 d) 4

Let g(x) be a function such that g(x+1)+g(x-1)=g(x) for every real x. Then for what value of p is the relation g(x+p)=g(x) necessarily true for every real x?

1) 5
2) 3
3) 2
4) 6
5) none of these

My take 6 ans

g(x+1)+g(x-1)=g(x) Means middle value is the sum of immediate left and right value.
in this process every value will repeat itself after 6 No's

EG. 6 7 1 -6 -7 -1 6 7 1 -6 -7 -1 6 7 1 -6 -7 -1 and so on

For g(x+p)=g(x) .

Now whatever the value of x , will give the same result on RHS after adding 6 as the value of p

Let g(x) be a function such that g(x+1)+g(x-1)=g(x) for every real x. Then for what value of p is the relation g(x+p)=g(x) necessarily true for every real x?

1) 5
2) 3
3) 2
4) 6
5) none of these

g(2)+g(0) = g(1)
g(3)+g(1) = g(2)
g(1) = g(2)-g(3)


=>g(2)-g(3) = g(2)+g(0)
=> -g(3)=g(0)
=>g(3) = -g(0)

in the same way
-g(6) = g(3)

=>-g(0) = -g(6)
g(0) = g(6)

so p = 6

lets try these question

Q.1 find n if 2 ^200 - ( 2^192 *31 ) + 2^n is a perfect square ?

Q .2 find out n for 3^9 + 3^12 + 3^15 + 3 ^n is a perfect cube ?

Q.3 find out remainder 32! / 37 ?

ans 2 :-

= > 3^9 + 3^12 + 3^15 + 3 ^n = 3^9 { 1+ 3^3 + 3^6 + 3 ^ n-9 }

3^9 {1 + 3* 3^2 + 3 * (3^2)^2 + 3^(n-9 ) - 3 (3^2)^2 }


so 3^(n-9 ) - 3 (3^2)^2 =0

or n-9 = 5

or n =14

ans 3:- 32!/37


35! / 37 = 1

so ( 35 * 34 * 33! ) / 37 = -36 (negative remainder )

6 *33! /37 = -36

33 ! /37 = -6

33 * 32! / 37 = 31

(-4) * 32! /37 = 31 or 37k+31

32! /37 = -17

or 32! / 37 = 20

so remainder 20

Let g(x) be a function such that g(x+1)+g(x-1)=g(x) for every real x. Then for what value of p is the relation g(x+p)=g(x) necessarily true for every real x?

1) 5
2) 3
3) 2
4) 6
5) none of these

If the HCF of any two numbers in set A is 3, then what is the maximum numbers of elements that set A can have. Set A is formed by selecting first 100 natural numbers.

1) 10
2) 12
3) 11
4) 14
5) 15

we have to take 3*p where p is 1 or a prime number

p can be 1 2 3 5 7 11 13 17 19 23 29 31

A = {3 6 9 15 21 33 39 51 57 69 87 93} => 12 numbers

My take 11

3,2*3,3*5,3*7,3*11,3*13,3*17,3*19,3*23,3*29,3*31

is it 12
3,3*2,3*5........3*31 and 3^2

sam05 Says

i think we can also take 9

Correct..........

12 is OA................

My take 11

3,2*3,3*5,3*7,3*11,3*13,3*17,3*19,3*23,3*29,3*31

i think we can also take 9

If the HCF of any two numbers in set A is 3, then what is the maximum numbers of elements that set A can have. Set A is formed by selecting first 100 natural numbers.

1) 10
2) 12
3) 11
4) 14
5) 15

is it 12
3,3*2,3*5........3*31 and 3^2

2-(a+b)^3 = a^3+3a^2b+3ab^2+b^3
here a = 3^5 and b = 3^3

so (3^5+3^3)^3 = 3^15+3^9+3^12+3^14

so n = 14


3-
32!/37
we now 35! mod 37 = 1
so 32!*33*34*35 mod 37 = 1
=> 32!*13 mod 37 = 1
=> 13*17 mod 37 = -1
=>13*-17 mod 37 = 1

so 32! mod 37 = -17 or 20

32! / 37 remainder = 20

If the HCF of any two numbers in set A is 3, then what is the maximum numbers of elements that set A can have. Set A is formed by selecting first 100 natural numbers.

1) 10
2) 12
3) 11
4) 14
5) 15

My take 12

3,2*3,3*5,3*7,3*11,3*13,3*17,3*19,3*23,3*29,3*31 & 9

sam05 Says

i think the third answer should be 20.its coming as -17 for me.

35! mod 37 = 1
32!*33*34*35 mod 37 = 1
32!*-4*-3*-2 mod 37 = 1
32!*-24 mod 37 = 1
32!*13 mod 37 = 1
17*13 mod 37 = -1
-17*13 mod 37 = 1

32! mod 37 = -17 or 20

yes it was mistake...:banghead: edited

2-(a+b)^3 = a^3+3a^2b+3ab^2+b^3
here a = 3^5 and b = 3^3

so (3^5+3^3)^3 = 3^15+3^9+3^12+3^14

so n = 14


3-
32!/37
we now 35! mod 37 = 1
so 32!*33*34*35 mod 37 = 1
=> 32!*13 mod 37 = 1
=> 13*17 mod 37 = 1

so 32! mod 37 = 17

=> 32!*13 mod 37 = 1
culdip bhai how to solve this line. plz reply

If the HCF of any two numbers in set A is 3, then what is the maximum numbers of elements that set A can have. Set A is formed by selecting first 100 natural numbers.

1) 10
2) 12
3) 11
4) 14
5) 15

lets try these question

Q.1 find n if 2 ^200 - ( 2^192 *31 ) + 2^n is a perfect square ?

Q .2 find out n for 3^9 + 3^12 + 3^15 + 3 ^n is a perfect cube ?

Q.3 find out remainder 32! / 37 ?

2-(a+b)^3 = a^3+3a^2b+3ab^2+b^3
here a = 3^5 and b = 3^3

so (3^5+3^3)^3 = 3^15+3^9+3^12+3^14

so n = 14


3-
32!/37
we now 35! mod 37 = 1
so 32!*33*34*35 mod 37 = 1
=> 32!*13 mod 37 = 1
=> 13*17 mod 37 = -1
=>13*-17 mod 37 = 1

so 32! mod 37 = -17 or 20

lets try some more question


determine the set of integers n for which n^2 +19 n + 92 is a square ?

let n^2 +19 n + 92 = m^2 (m is a non negative integer )

then n^2 +19 n + 92 - m^2 = 0

solve for n n = 1/2 { -19 + or - _/(4m^2-7) }

so 4m^2-7 is a square 4m^2 - 7 = p^2

so
( 2m - p ) ( 2m +p ) = 7 so

m and p are natural no so

2m - p = 1

and 2m +p = 7 hence m = 2

so

n^2 +19 n + 92 = 4
n^2 +19 n + 88 = 0
n = -8 or -11

certain shade of gray paint is obtained by mixing 3 parts of white paint with 5 parts of black paint. If 2 gallons of the mixture is needed and the individual colors can be purchased only in one-gallon or half- gallon cans, what is the least amount of paint, in gallons, that must be purchased in order to measure out the portions needed for the mixture?
(a) 2 (b) 2 .5 3 (d) 3.5 (e) 4

koyi kar do bhai ......!!!!

3/3= 1
5/3 =~ 1.5

2.5

3/3 =1
5/3 =~ 1.5(assuming the higher value)

2.5?

Certain shade of gray paint is obtained by mixing 3 parts of white paint with 5 parts of black paint. If 2 gallons of the mixture is needed and the individual colors can be purchased only in one-gallon or half- gallon cans, what is the least amount of paint, in gallons, that must be purchased in order to measure out the portions needed for the mixture?
(A) 2 (B) 2 .5 3 (D) 3.5 (E) 4

koyi kar do bhai ......!!!!

in 2 gallon gray paint there will be 1.25 black and .75 white paint

but as paint can be purchased in 1 or 1/2 gallon cans
so we have to purchase 1.5 gallon black and 1 gallon white paint
total 2.5 gallon

Certain shade of gray paint is obtained by mixing 3 parts of white paint with 5 parts of black paint. If 2 gallons of the mixture is needed and the individual colors can be purchased only in one-gallon or half- gallon cans, what is the least amount of paint, in gallons, that must be purchased in order to measure out the portions needed for the mixture?
(A) 2 (B) 2 .5 3 (D) 3.5 (E) 4

koyi kar do bhai ......!!!!

( 3/8 )*2 = 3/4 gallon but cans are available in half or full gallon


so 1 gallon required


and for black ( 5/8 ) * 2 i.e. 5/4 required
but available can in half or full gallon

so 3/2 are required

so total 1 + 1.5 = 2.5 gallon are required

lets try these question

Q.1 find n if 2 ^200 - ( 2^192 *31 ) + 2^n is a perfect square ?

Q .2 find out n for 3^9 + 3^12 + 3^15 + 3 ^n is a perfect cube ?

Q.3 find out remainder 32! / 37 ?

deleted post

Certain shade of gray paint is obtained by mixing 3 parts of white paint with 5 parts of black paint. If 2 gallons of the mixture is needed and the individual colors can be purchased only in one-gallon or half- gallon cans, what is the least amount of paint, in gallons, that must be purchased in order to measure out the portions needed for the mixture?
(A) 2 (B) 2 .5 3 (D) 3.5 (E) 4

koyi kar do bhai ......!!!!

bs0409,if p,q are non zero elements

its 11
5(p+q)(p-q)+11pq would be minimum when p=q=1.

lets try some more question


determine the set of integers n for which n^2 +19 n + 92 is a square ?

n^2 +19 n + 92 = (n+19/2)^2 +7/4
=> (2n+19)^2+7 = k^2
=> k^2-(2n+19)^2 = 7
=> (k+2n+19)(k-2n-19) = 7 = 1*7 or -1*-7

k+2n+19 = 7 and k-2n-19 = 1
k = 4 , n = -8

or
k+2n+19 = 1 and k-2n-19 = 7
k = 4 , n = -11

or k+2n+19 = -1 and k-2n-19 = -7
k = -4 and n = -8

or k+2n+19 = -7 and k-2n-19 = -1
k = -4 and n = -11

so n can be -8 or -11 => 2 solutions

ANS=23..........

Birthday problem - Wikipedia, the free encyclopedia

Thnx for sharing the link its very useful

is it 11??

Plzzz help me out

There are N people in one room. How big does N have to be until the probability that at least two people in the room have the same birthday is greater than 50 percent?

ANS=23..........

Birthday problem - Wikipedia, the free encyclopedia

Plzzz help me out

There are N people in one room. How big does N have to be until the probability that at least two people in the room have the same birthday is greater than 50 percent?

find the remainder of 3^1994/1000
how to go abt this???

ANS=369

3^1994=9^997=(10-1)^997=1000k-997C995*100+997C996*10-1
=0-600+970-1 (mod 1000)
=369

lets try some more question


determine the set of integers n for which n^2 +19 n + 92 is a square ?

bs0409,if p,q are non zero elements

In the given equation 5p^2+11pq-5q^2 the term which can decrease our total value is -5q^2.
Hence whenever p=q only middle term contributes.
So min value of the quation will be when p=q=1, Hence min value of the given quation will be 11

sam05 Says

exactly.but to know that 729*369 is of that form the options are very useful

ys optn can be useful bt without optn we ca do it too..
see.. we have to make last 3 digit 001..
so 729 *9 will have last digit 1 ...
729*9= 6561 .......(1)
nw we have to make that 6 0
so 729*60 = 43740..... (2)

add 1 nd 2 ...xxxx301 ... to make that 3 0
729*300 = xxxxx700 ....(3)
adding ol u get last 3 digit 001

3^1994 mod 1000
1000 = 8*125
remainder by 8
3^2 mod 8 = 1
so 3^1994 mod 8 = 1

now
3^1994 mod 125
euler of 125 is 100
so 3^2000 mod 125 = 1

suppose 3^1994 mod 125 = k
k*3^6 mod 125 = 1
k*4 mod 125 = 1
-31*4 mod 125 = -1
so 94*4 mod 125 = 1
so k = 94
or 3^1994 mod 125 = 1

remainder by 1000 will be in the form of 125k+94 and 8m+1
or 5k+5 is divisible by 8
least value of k is 7

so remainder will be 125*7+94 = 969

but sir from where are you getting this 5k+5 divisible by 8 thing????

what is the minimum value of 5p^2+11pq-5q^2,if p,q are non zero elements

e(1000) = 400
now 3^2000 wl leave remainder 1 when divided by 1000
so 1 = 3^6 * 3^1994
= 729 *3^1994
now to make remainder 1 last 3 digit of 729*3^1994 has to be 001
so
729*369= xxx0001 so remainder has to be 369

exactly.but to know that 729*369 is of that form the options are very useful

find the remainder of 3^1994/1000
how to go abt this???

3^1994 mod 1000
1000 = 8*125
remainder by 8
3^2 mod 8 = 1
so 3^1994 mod 8 = 1

now
3^1994 mod 125
euler of 125 is 100
so 3^2000 mod 125 = 1

suppose 3^1994 mod 125 = k
k*3^6 mod 125 = 1
k*104 mod 125 = 1
6*104 mod 125 = -1
so -6*104 mod 125 = 1
so k = -6

remainder by 1000 will be in the form of 125n-6 and 8m+1
or 5n-7 is divisible by 8
least value of k is 3

so remainder will be 125*3-6 = 369

find the remainder of 3^1994/1000
how to go abt this???

e(1000) = 400
now 3^2000 wl leave remainder 1 when divided by 1000
so 1 = 3^6 * 3^1994
= 729 *3^1994
now to make remainder 1 last 3 digit of 729*3^1994 has to be 001
so
729*369= xxx0001 so remainder has to be 369

3 ^ 1994 /1000

apply eulers method

#(1000) = 1000 * (1 - 1/2) (1 - 1/5)
= 400


3 ^ 2000 / 1000 = gives remainder 1

3 * 3 ^ 1999 /1000 = -999 ( negative remainder )
3 ^1999 /1000 = -333
3^2 * 3^ 1997 / 1000 = -333
3 ^1997 / 1000 = -37

3 ^ 1997 /1000 = 963

3 ^2 * 3^1995 /1000 = 9 * 107

3^1995 /1000 = 107

3 * 3^1994 = 1000k+107

so put k =1

3 * 3^1994 = 1107
3 * 3^1994 = 3* 369

so 3^1994 /1000 = remainder 369

too good.actually even i was trying with euler.but got stuck in middle cause didn have options.with options the sum would have been very easy.but superb method sir.

Bababhokali Says

That's right

I think its arun sharma question

ajeetaryans Says

Let a,b,c,d,e be consecutive positive integers such that b+c+d is a perfect square and a+b+c+d+e ia perfect cube . find smallest possible value of c

a,b,c,d,e are consecutive positive integers
so b + c+ d = 3c

and

a + b + c + d + e = 5c

now 3c is a square so 3 c = 9 *k^2

and 5c is a cube so 5c = 125 * m^3


so c = 3*k^2 = 5^2 * m^3

so least possible value of c

3 * 3^2 * 5^2

= 675

Number of days be d and work per day be x

xd=1800

(x-20)d/3+(x+20)2d/3=1800+x+20
20d=3x+60
36000=3x^2+60x
Solving the above equation we get
x=-120 or x=100
So x=100 tonnes per day

That's right

a team of miners planed to mine 1800 tons of ore durinng a ceratain number of days.due to technical problem in one third of the planned number of days,team was able to achieve an output of 20 tons of ore less than the planned output,to make for this,team over achieved for rest of days by 20 tons.result was that team completed the task one day ahead of time.how many tons of ore did the team intitaly plan to ore per day.

Number of days be d and work per day be x

xd=1800

(x-20)d/3+(x+20)2d/3=1800+x+20
20d=3x+60
36000=3x^2+60x
Solving the above equation we get
x=-120 or x=100
So x=100 tonnes per day

find the remainder of 3^1994/1000
how to go abt this???

3 ^ 1994 /1000

apply eulers method

#(1000) = 1000 * (1 - 1/2) (1 - 1/5)
= 400


3 ^ 2000 / 1000 = gives remainder 1

3 * 3 ^ 1999 /1000 = -999 ( negative remainder )
3 ^1999 /1000 = -333
3^2 * 3^ 1997 / 1000 = -333
3 ^1997 / 1000 = -37

3 ^ 1997 /1000 = 963

3 ^2 * 3^1995 /1000 = 9 * 107

3^1995 /1000 = 107

3 * 3^1994 = 1000k+107

so put k =1

3 * 3^1994 = 1107
3 * 3^1994 = 3* 369

so 3^1994 /1000 = remainder 369

find the remainder of 3^1994/1000
how to go abt this???

Getting ans as 29..What's OA??