Let f(x,y) be defined as the remainder when (x–y)! is divided by x. If

Question

Let f(x,y) be defined as the remainder when (x–y)! is divided by x. If x=25, what is the maximum value of y for which f(x,y)=0

A. 5
B. 10
C. 15
D. 20
E. 25

manpreetsingh86 · Answer

we have to find the value of y for which (25-y)!/ 25 yields zero remainder

if y=20, then (25-20)!=5!, which clearly is not divisible by 5^2
thus our answer will be less than 20

let's try 15 (25-15)!=10!, which is clearly divisible by 5^2.

hence C

Bhanupriyachauhan · Answer

Answer - D 
1/25 no reminder

So simple

(X-y)!/x = no reminder
If x = 25 
Then (25-x)!/25 = no reminder
Let y = 25 
Then 0!/25 , 0! = 1 
So 1/25 which have 0 reminder
So answer D

Bunuel · Answer

That's not true. 1 divided by 25 yields the remainder of 1, not 0.

Bhanupriyachauhan · Answer

Totally agree. 
It was my mistake.
Sorry

Posted from my mobile device

LighthousePrep · Answer

It's important to note why 10! is clearly divisible by 5^2, while 5! is not.

It is possible to find the prime factors and their powers of a factorial by counting the amount of times a prime factor p is contained within some number n for n!.

This is most easily done by taking the floor value, or rounding to the next highest whole integer, when dividing out n by p to see how many numbers between p and n have a factor of p.
One must also remember to be mindful of the fact that there most likely exist numbers with multiple factors of p, in which case you must keep dividing by p until floor(p) = 0.

In this case, 5! clearly has no factors of 5^2 because floor(5/5) (or n/p, where n is taken from 5! and p is taken from 5^2) = 1, and floor(1/5) = 0. This means that the power of 5 in the prime factorization of 5! is 1, so it does not contain 5^2.

10!, however, clearly has a factor of 5^2 because floor(10/5) = 2, so the prime factorization of 10! contains 5^2. Continuing this factorization further purely to see how this method works, the prime factorization contains 8 2's (floor(10/2) = 5, floor(5/2) = 2, floor(2/2) = 1, 5+2+1 = 8).

I hope this method can help to quickly identify prime factors of factorials!

Temurkhon · Answer

The question is in finding  y  to be (25-y)! multiple of 25. That means we should have at least 5*5 in (25-y)!

Go directily answer choices and start with 15, meaning we have 10!, having 5*5. But we should get maximum for  y , so go 20, meaning we have 5!, not having 5*5.

C

Beat720 · Answer

(x-y)! = x*n + f(x,y)
f(x,y) = 0; x=25 --> (25-y)! = 25*n --> (25-y)! is divisible by 5*5. Among options, D and E make (25-y)! not divisible by 5*5: eliminate
Find Max y = find Min (25-y) --> among A, B, C, C offers the min (25-y) and max y. --> C

rever08 · Answer

Bunuel  why isn't it E? Please help!

soodia · Answer

I have the same question, 0/25.... reminder will be 0..
is it wrong?

Bunuel · Answer

f(x,y) is the remainder when (x–y)! is divided by x;

If x = y = 25, then f(x,y) is the remainder when (25 - 25)! = 0! = 1 divided by 25, so f(x,y) = f(25, 25) = 1, NOT 0.

KarishmaB · Answer

f(x, y) = Remainder of  \frac{(x - y)!}{x}

For 0 remainder, (x - y)! should have x (=25) as a factor. So (25 - y)! should have at least two 5s.

Note that the first factorial to have two 5s is 10!.
So y can be at most 15 to give (25 - 15)! = 10!
This will be divisible by 25.

Answer (C)

ssrovi · Answer

f(x,y)=(x-y)!/x

since x= 25
we need atleast two multiples of 5 in (x-y)! i.e. 5,10 which is possible in 10!
so x-y=10
y=x-10=25-10=15
Max value of y is 15

Andrewcoleman · Answer

zero factorial equals 1 not 0

0!=1

udaypratapsingh99 · Answer

For (25-y)! to be divisible by 25, (25-y)! must have 2 5s in it, 
ie (25-y)/5 >=2
=> y =< 15

C is Correct

sjmonroe · Answer

Why can't y=25? Is the remainder of (25-25)!/25 not zero?

Bunuel · Answer

Please read the thread: https://gmatclub.com/forum/let-f-x-y-be ... l#p1963001

Let f(x,y) be defined as the remainder when (x–y)! is divided by x. If

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