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

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

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

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

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.

­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!