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Bunuel
What is the greatest integer p, for which 3^p is a factor of 21! ?

(A) 8
(B) 9
(C) 10
(D) 11
(E) 12

To determine the number of factors of 3 within 21!, we can use the following shortcut in which we divide 21 by 3, and then divide the quotient of 21/3 by 3 and continue this process until we can no longer get a nonzero integer as the quotient.

21/3 = 7

7/3 = 2 (we can ignore the remainder)

Since 2/3 does not produce a nonzero quotient, we can stop.

The final step is to add up our quotients; that sum represents the number of factors of 3 within 21!.

Thus, there are 7 + 2 = 9 factors of 3 within 21!.

Answer: B
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Solution:



Given:

    • \(p\) is an integer

    • \(3^p\) is a factor of \(21!\)

Working out:

We need to find the greatest value of “p

Since \(3^p\)is a factor of \(21!\), and we need to find out the greatest integer p, it can be inferred that 3^(p+1) is not a factor of p.

    • We need to calculate the number of 3s in 21!
Since all the numbers in 21! is in the multiplicative form, the powers of 3 will be added.

    • Multiples of 3 in 21! are: 3, 6, 9, 12, 15, 18, 21.

      o All the above 7 numbers have at least one 3 in them. However, 9 and 18 contain two 3s in them.

      o Thus, the total number of 3s in 21! = 7+2 =9

Hence \(3^9\) is a factor of \(21!\) but \(3^{10}\) is not.

Thus, the greatest integer value of p is 9

Answer: Option B
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To find the highest power of a prime number (x) in a factorial (N!), continuously divide N by x and add all the quotients.

Here it is calculated as 21/3 +7/3 +2/3 = 7+ 2+0 = 9.
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To find the highest power of a prime number (x) in a factorial (N!), continuously divide N by x and add all the quotients.

Here it is calculated as 21/3 +7/3 +2/3 = 7+ 2+0 = 9.
Nice approach


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