If N is the product of all positive integers less than 31,

Question

If N is the product of all positive integers less than 31, than what is the greatest integer k for which N/18^k is an integer?

A. 3
B. 6
C. 7
D. 14
E. 26

Bunuel · Accepted Answer

If N is the product of all positive integers less than 31, than what is the greatest integer k for which N/18^k is an integer?

A. 3
B. 6
C. 7
D. 14
E. 26

Check this: everything-about-factorials-on-the-gmat-85592.html

Given: \(n=30!\). Question: if \(\frac{30!}{18^k}=integer\) then \(k_{max}=?\)

We should determine the highest power of 18 in 30!.

\(18=2*3^2\), so we should find the highest powers of 2 and 3 in 30!:

Highest power of 2 in 30!: \(\frac{30}{2}+\frac{30}{4}+\frac{30}{8}+\frac{30}{16}=15+7+3+1=26\), --> \(2^{26}\);

Highest power of 3 in 30!: \(\frac{30}{3}+\frac{30}{9}+\frac{30}{27}=10+3+1=14\) --> \(3^{14}\);

\(n=30!=2^{26}*3^{14}*p\), where \(p\) is the product of other multiples of 30! (other than 2 and 3) --> \(n=30!=(2*3^{2})^7*2^{19}*p=18^7*2^{19}*p\) --> so the highest power of 18 in 30! is 7 --> \(\frac{30!}{18^k}=\frac{18^7*2^{19}*p}{18^k}=integer\) --> \(k=7\).

Answer: C.

Hope it's clear.

jlgdr · Answer

Hi all, this was my approach for solving this.

Basically the question is asking us what is the highest power of 18 that can give us a number that is a factor of 31!

So remember 18 after prime factorization is (3^2)(2). Now there are going to be less factors of 3 in 31!, than factors of 2.
Therefore, lets find how many factors of 3 in 31! We can use this quick method

31/3^1 = 10
31/3^2= 3
31/3^3=1
Sum = 14
Just ignore the remainders.

So we have that 3^14 must be the least. Now don't forget that 18 is 3^2k so k must be ONLY 7, because 2k will give us the 14.
Hence answer is (C)

Hope it helps

Bunuel could you please validate this one? Thank you
Cheers
J

madn800 · Answer

According to Bunuel  18=2*3^2

I listed 14 3's
14 3's-  3 3   3 3   3 3   3 3   3 3  3 3 3 3
 
26 2's-  2   2   2   2   2  2 2 2 2 2 2 2 2 2 and so on

notice that 7 such combinations of 18 are possible.
so answer=  18^7

KarishmaB · Answer

Responding to a pm: Both the methods are the same. In some cases, you need to work on more than one factor, in others you don't. Here is how we decide: - If each factor appears only once in the number, you just need to find the number of times the largest factor appears. The largest factor will have the lowest exponent. e.g. 6 (2*3), 10 (2*5), 30 (2*3*5) etc - If the larger factor appears multiple times in the number, again you just need to find the number of times the largest factor appears. e.g. 18 (2 * 3 *3), 50 (2 * 5 * 5) etc - If the smaller factor appears more than the greater factor in the number, then you need to work on both factors. e.g. 12 (2 * 2 * 3), 175 (5 * 5 * 7) etc In this case, the smaller factor appears more often that the greater factor in the factorial but you actually need more of each smaller factor to make each number. So you don't know what will dominate. For more on this, check out: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/06 ... actorials/ The post as well as the comments.

amitsah · Answer

here n=30!
now we have to find highest power of 18 in 30!. this is done in this way
 18 = 2*3^2
so, we find the highest power of 2 and 3^2 in 30! individually 
 
for 2

30/2 + 30/4 + 30/8 + 30/16 
 15+7+3+1
 26

similarly for 3
30/3+ 30/9+30/27
10+3+1
14
so, there are 14/2= 7, power of 3^2 in 30! 
so required answer is 7 as we need seven (3^2) and seven (2) to make their products as power of 18.

JeffTargetTestPrep · Answer

We see that N = 30! and know that k is an integer.
 
N/18^k = N/ ^k = N/ 
 
The expression above is an integer if k is not greater than the total number of 2s in the prime factored form of 30! and if 2k is not greater than the total number of 3s in the prime factored form of 30!
 
Since in the prime factored form of 30!, there are fewer 3s than 2s, we need to concentrate on the total number of 3s, which we can quickly determine with the following technique:
 
30/3 = 10
 
10/3 = 3 (ignore the remainder)
 
3/3 = 1
 
The total number of 3s is 10 + 3 + 1 = 14, so it must be true that:
 
2k ≤ 14
 
k ≤ 7
 
We see that the maximum value of integer k is 7.
 
 Answer: C

Akriti_Khetawat · Answer

I have a doubt here, what do we do with the 26 that we got? That's the number of 2's in the number right? Here we've divided only 14 by 2 to get the answer. Please help!

Bunuel · Answer

18 = 2*3^2.

30! has 2 to the power of 26 and 3 to the power of 14. However, we need 3^2, not simply 3, and 3^2 in 30! is to the power of 14/2 = 7.

Checke detailed explanation here: https://gmatclub.com/forum/if-n-is-the-product-of-all-positive-integers-less-than-103218.html#p803385

bumpbot · Answer

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If N is the product of all positive integers less than 31,

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