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

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

Totally agree.
It was my mistake.
Sorry

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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!

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

(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

Bunuel why isn't it E? Please help!

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

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.

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)

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

zero factorial equals 1 not 0

0!=1

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