If p is the product of the integers from 1 to 30, inclusive, what is

Concept: Highest power on x in n! :
Highest power of x(x^p) that will divide n! is p = \(x/n + x/n^2 + x/n^3 ........\)

To make it easy here they are asking for highest power of 3 in 30!
So 30/3 =10,
10/3 = 3,
3/3 = 1.
So p = 10+3+1 = 14.

Tip : Instead of dividing n with \(x,x^2,x^3\) separately we can divide the resultant as shown above.

To find the greatest integer k for which 3^k is a factor of p, we need to determine how many times 3 appears as a factor in the product of the integers from 1 to 30.

Since we are looking for the number of times 3 appears as a factor, we can count the number of multiples of 3 among the integers from 1 to 30.

There are 10 multiples of 3 between 1 and 30, namely: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30.

However, among these multiples, there are additional factors of 3 in numbers such as 9 and 27, which have multiple factors of 3.

So, let's count the total number of factors of 3 among the multiples of 3 between 1 and 30:

3 (1 factor of 3)
6 (1 factor of 3)
9 (2 factors of 3)
12 (1 factor of 3)
15 (1 factor of 3)
18 (2 factors of 3)
21 (1 factor of 3)
24 (1 factor of 3)
27 (3 factors of 3)
30 (1 factor of 3)

Adding up the total number of factors of 3, we have 1+1+2+1+1+2+1+1+3+1=14.

Therefore, the greatest integer k for which 3^k is a factor of p is k=14.

The correct answer is (C) 14.

Product of the integers from \(1\) to \(30\) = \(p = 1*2*3*...............27*28*29*30 = 30!\)

Greatest integer \(k\) for which \(3^k\) is a factor of \(p\)

The question boils down to highest power of \(3^k\) in \(30!\)

So, the value is

\(\frac{30}{3} = 10\)
\(\frac{10}{3} = 3\)
\(\frac{3}{3} = 1\)

Hence, the value of \(k\) is \(10 + 3 + 1 = 14\), Answer must be (C)

If p is the product of the integers from 1 to 30, inclusive, what is the greatest integer k for which 3^k is a factor of p?

A. 10
B. 12
C. 14
D. 16
E. 18

P= 30!. Multiples of 3 which single powers and double powers sum upto 14. Hence C

I tried this for 25!, and divided it by 2... couldn't get the correct answer (got 24, after considering the reminders)... can some try it and repost

The answer is simply dividing continuously by the number whose power you require.
25! by 2 means you are looking for value of k in 2^k

Then 25/2 + 12/2 + 6/2 + 3/2 = 12+6+3+1 = 22

Check out this post: https://anaprep.com/number-properties-h ... actorials/

­
The easiest  solution to this: Hope this helps. 

­