For nonnegative integers x , y , and m , what is the greatest value

Hi,
A Good Q..
the Q statement is "For nonnegative integers x, y, and m, what is the greatest value of m for which x^m is a factor of y!?"
What doe sthis mean..
It means the largest power of x that is there in y!..
formula is \(\frac{y}{x} + \frac{y}{x^2}..\) and so on till the fraction \(\frac{y}{x^z}\) becomes less than 1..
x is a prime number or the biggest prime in any integer..

lets see the sentences..
(1) y=x−1
this means \(x^m\) in (x-1)!..
\(\frac{y}{x} + \frac{y}{x^2}..\). and so on means \(\frac{(x-1)!}{x}\)..
if x is prime, answer is 0..
if not it will depend on x..
say x=6, so y=5..
check for 3s in 5! as 3 is the largest prime number in 6..
\(\frac{5}{3}\)=1 so m=1..
different answers
But we do not know if x is prime or what is the largest prime in the integer x..
insuff


(2) x is a prime number
since there is no corelation in y and x, we cannot answer ..
say y is 50 and prime is 5, then it is \(\frac{50}{5}+\frac{50}{25}=12.\).
and say 4 and prime is 3then 4/3=1..
insuff..

combined .
we know that x is prime and y is x-1..
from this it becomes clear that y! or (x-1)! will not have x..
so power of x will be 0, or m=0..
suff
C
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We need the greatest value of m for which x^m is a factor of y!
We have no relative values of x, y and m given in the question stem.

(1) y=x−1

Think what this means. Say x = 8. Then y = 7
We need greatest m such that 8^m is a factor of 7!
7! has a 2, a 4 and a 6 so it has four 2s. So it can make one 8. So m = 1.
But what if x = 7? Then y = 6
We need greatest m such that 7^m is a factor of 6!. Can we make 7 out of first 6 numbers? No, because 7 is prime. So it means that you can never make 7 out of any other 2 factors. So m = 0
Not sufficient.

(2) x is a prime number
x could be 7 but y could be either 6 or 10 (or infinite other values). 6! will not have any 7s but 10! will.
Not sufficient.

Using both:
This is our second case above. If x = prime, y is prime - 1. All numbers till prime - 1 will not be able to make a single prime.
i.e. our second case above. If x = 7, then y = 6 and all positive integers till 6 will not be able to make a 7. So m will always be 0.

Answer (C)
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Aside: Whenever you see the word "non-negative," you should ask yourself "Why didn't they just say POSITIVE?"
The answer to this question is a big hint.
Non-negative is not the same as positive, since ZERO is a non-negative number and zero is not positive.
So, when you see the word "non-negative," be sure to consider the possibility that the number equals zero.

Cheers,
Brent

This took me 4 minutes to solve
Any other methods?

Is there any alternate way to crack this problem. Bunuel please help. I tried solving it by making cases and took somewhere close to 3 minutes to figure it out.

This took me only 1 min to solve.

Here is how I approached.

We are given x, y, and m and non negative integers. So, we need to have x,y and m >=0;

We need to find the greatest value of m such that y!/x^m = Integer.

Now Statement 1 : y=x−1. It means x and y are co-primes. Now, I checked the fraction for x=2, y=1, for this m has to be zero. Now I checked for a larger number say x=10; y = 9, for this m could be 0 and 1 but not greater than that. -- INSUFFICIENT.

Statement 2 : x is prime. Nothing about y. Insufficient

Combining : So, for any value of prime, we will always have m = 0 because positive value of m will NEVER make the two co primes divisible.. HENCE I MARKED C.
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Just ensuring that my understanding of the concepts is correct.
We have to find the maximum value of m for which x^m is is a factor of y!

Statement 1
y+1=x
now if y=5 x=4 then m=1
but if
y=10 and x=9 them m=2
NS

Statement 2 x is prime quite not sufficient in itself.

However combining 1 and 2 we have the following situation where x is prime and y is a coprime thus the only factor is 1
so C

I guess I am making a mistake here. Please correct me

Question asks: how many number x can be formed using elements in y!.
Stat.1 gives x=y+1. So x is not in y!. Can x be constructed from elements of y!? Stat.2 gives that x is prime, thus x can't be built from other numbers. Answer: C.

Very tricky.

1) You should see X could be 1 and Y is 0!, 0! is 1 (for whatever reason) therefore M could be any number. Also if X=2, Y=1 therefore M =0 So not sufficient
2) If X =2, Y has no restriction. So Y! could also be 2! so M=1. But Y could also be 9! and M could equal 2 as 3^2 is a factor or 9! as 9! contains a 9 which is a factor of 3^2. So this number could be infinite.

In terms of C--We know Y! is always less than X and X is prime. So if Y! is less than X we know the prime is always above 6! A prime number such as 7^M doesn't go into 6! Because 7 is a prime and there are no prime factors of 7 in 6X5X4X3X2X1. Therefore, M always has to be zero because only 1 is a factor. This is now sufficient.

Using 1: Possible values of m:
x : 1 2 3 4 5 6 7 8 9 10 ..
y : 0 1 2 3 4 5 6 7 8 9 ..
y! : 1 1 2 6 24 120 720 ..
m : 0 1 0 0 0 1 0 1 1 1..

Also, as x and y are co-prime, I see 'm' takes only two values 0 or 1
We can see the 'greatest value' of m for which x^m is a factor of y! is '1'
Thus, A is sufficient

VeritasKarishma, BrentGMATPrepNow, chetan2u: Can you please point the error in my reasoning?
Thanks.

chetan2u

What if when X = 2. Now 2 is a prime number and for 2^m to be divisible by (2-1)! or 1!, m can take any value. How is C sufficient ?

You are taking the opposite of what is given.
x^m or 2^m has to be factor of 1!
2^m will be a factor of 1! only when m =0 as 2^0=1.
But if m is 1, 2^1 is 2 and will not be a factor of 1!
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Hi chetan2u

Can you please derive how the above formula is linked to this ques? and how can we infer that "x is a prime number or the biggest prime in any integer."?

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