TomB wrote:

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

Responding to a pm:

**Quote:**

If 6^y is a factor of (10!)^2, What is the greatest possible value of y ?

The maximum powers of a prime number 3, in 10!: 103+1032=3+1=4103+1032=3+1=4 (take only quotients into account). So, we have that the maximum power of 3 in 10! is 4, thus maximum power of 3 in (10!)^2 will be 8: (34)2=38(34)2=38. As discussed 8 is the maximum power of 6 as well.

But the below question we are taking both the factors 2 and 3 .Please can you explain the differnece

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? we are factoring 18 as 2^2 and 3.

Given: n=30! Question: if 30!/18k=integer kmax=?

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!: 30/2+30/4+30/8+30/16=15+7+3+1=26, --> 2^26

Highest power of 3 in 30!: 30/3+30/9+30/27=10+3+1=14 --> 3^14

n=30!=2^26∗3^14∗ where p is the product of other multiples of 30! (other than 2 and 3) --> n=30!=(2∗3^2)^7∗2^19∗p--> so the highest power of 18 in 30! is 7

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:

http://www.veritasprep.com/blog/2011/06 ... actorials/The post as well as the comments.

okay with first and second scenario ....

For the third scenario, however, say, 10!/12...

here, we have 2 2s and 1 3...so the smaller factor appears more often than does the bigger factor...

the highest power 12 can have is 4.....

please correct if I am missing something, mam ...