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

Question

For nonnegative integers x, y, and m, what is the greatest value of m for which x^m is a factor of y!?

(1) y=x−1

(2) x is a prime number

chetan2u · Answer

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

BrentGMATPrepNow · Answer

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

stonecold · Answer

This took me 4 minutes to solve
Any other methods?

keats · Answer

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.

KarishmaB · Answer

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)

abhimahna · Answer

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.

AmritaSarkar89 · Answer

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

plalud · Answer

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.

Samuelboyle96 · Answer

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.

CalvinHobbesIn · Answer

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.

Namangupta1997 · Answer

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 ?

chetan2u · Answer

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!

Kushchokhani · Answer

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."?

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

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

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards