Sorry sorry
same does not hold for 6 or 7.
You have to again do comparisions
thanx for the reply but what about the coins question...
can anyone plz help me to find the answer to the following to questions:
1.Arun Bikas and Chetakar have a total of 80 coins among them. Arun triples the number of coins with the others by giving them some coins from his own collection. Next Bikas repeats the same process. After this Bikas now has 20 coins. Find the number of coins he had at the beginning?
a. 11
b. 10
c. 9
d. 12
let no of coins wid
arun=x
bikas=y
chetakar=z
after 1st operation
arun=x-2y-2z
bikas=3y
chetkar=3z
after 2nd operation
arun=3x-6y-6z
bikas=7y-2x-2z
chetkar=9z
now,we get 2 equations
7y-2x-2z=20
x+y+z=80
solving we get y=20
so,20 coins is the answer.
Hope it's clear
Need help with this one
Given (2^32)+1 is exactly divisible by a certain number. Which of the following is also divisible by the same number ?
a: (2^96)+1
b: (2^16)-1
c: (2^16)+1
d: 7 X (2^33)
e: (2^64)+1
Need help with this one
Given (2^32)+1 is exactly divisible by a certain number. Which of the following is also divisible by the same number ?
a: (2^96)+1
b: (2^16)-1
c: (2^16)+1
d: 7 X (2^33)
e: (2^64)+1
Is it a: (2^96)+1
Is it a: (2^96)+1
yep. Kindly explain or hint
internet.fanboy Saysyep. Kindly explain or hint
Assume that the number is 'x'
so k*x = (2^32)+1
=> k*x - 1 = (2^32)
cube both the sides
=> (kx)^3 - 3(kx)^2 + 3(kx) - 1 = 2^96
=> (kx)^3 - 3(kx)^2 + 3(kx) = 2^96 + 1
=> x = 2^96 + 1
so (2^96)+1 is divisible by 'x'
- I went by observing the options... dunno if there is any other direct way...
what is the sum and product of factors of number 1000...???
find the remainder when (38^16!)^1777 is divided by 17
a. 1
b. 16
c. 8
d. 13
My answer 'a'. is it corct....??
topper@IITK Sayswhat is the sum and product of factors of number 1000...???
for such questions ,GOD has made formulae
if N= a^p*b^q*c^r.....whre a,b,c..are the prime factors of the number N
then sum of factors is given by :
*......
and the product of factors is given by :
N^(1/2)*
internet.fanboy SaysAns. 6 (right ?)
6 zeros
Perhaps this year dream wll get true iiiiiiiiiiiiiiiiiiiiiiimmmmmmmmm-a,b,c
no it is not 3^33^3
3^333
b coz 3^33^3 is 1.717 * 10^47
and 3^100 is 5.15 * 10^47 so power still remaining.......
tell me if i am wrong....
i think it should be
3^3^33.
Correct me,if wrong
i think it shld be
no it is not 3^33^3
3^333
b coz 3^33^3 is 1.717 * 10^47
and 3^100 is 5.15 * 10^47 so power still remaining.......
tell me if i am wrong....
Q.1 I have a drawing room of 12*10. I want to place tiles on the floor and I do not have monetary problem so tiles can be maximum. So how many tiles can be place.
There should be some more constraints here as i can place a single tile of 12*10 and I can also place infinite tiles.
If the tiles are square (as a constraint), then the minimum number of square tile is of the size 2*2 (HCF of 12 and 10). The number of tiles = 12*10/2*2 = 30 (Thats the minimum).
Q.2 13^17+14^17-15^17/ 17 what is the remainder.
The remainder should be 13 + 14 - 15 = 12
Q.3 if N^3/7, N^2/5 and N^5/11 so what are possible remainders of all cases.
Need options.
Q.4 (N^p - N)/ P so the number can be divided by instead of P where P is a prime number.
Didn't understand the question at all.:-(
Q.2 is correct=12 and In Q.1 maxmum number of tiles ...so now u can think ......if u need explntion i'll give you.
Q.3 here are the options
A: (0,1,5)(0,1,3)(0,1,7)
B: (2,5,7)(3,6,7 )(4,7,9)
C: (0,1,6)(0,1,4)(0,1,10)
D: (1,2,7)(1,3,5)(1,4,11)
Q.4 instead of P (prime number), the number can be divided by....??
can some one tell to find the remainder when divisor is composite number like:
Find remainder when 2^1990 is divided by 1990 ?
Remainder = 2
ernitin SaysRemainder = 2
explaination plz
topper@IITK Says25! has how many trailing 0's..?
25/5=5
5/5=1
so 5+1=6.
can some one tell to find the remainder when divisor is composite number like:
Find remainder when 2^1990 is divided by 1990 ?
solve this using chinese remainder theorem
1990=199x5x2
remainder of 2^1990/199 is 29 =r1
remainder of 2^1990/5 is 4
also since 2^1990/5 is 4 and 2^1990/2 is 0, you get 2^1990/10=4=r2
from chinese theorem 199x+10y=1 so x=-1 and y=20
also axr2+byr1=199x-1x4+10x20x29 = 5004
5004/1990=1024
So 1024 is the remainder
_______________snipped__________________
Two four digit numbers are written side by side to form an eight digit number which is divisible by the product of the two nos. The pair is
a) 1234 & 5678 b) 5000 & 2000 c) 1334 & 3667 d) 3334 & 1667
a) 1234 & 5678 b) 5000 & 2000 c) 1334 & 3667 d) 3334 & 1667