What is the greatest power that 5 can be raised to so that the resulti

There are only 3 multiples of 5 in 15!. Namely, 5, 10, and 15.
Each of these three contain 5 only once as a factor: 5 *1, 5*2 , 5*3
Hence, the greatest power of 5 is 3.

Yes you are correct. You will continue increasing power and dividing till the value after division becomes less than 1

VERITAS PREP OFFICIAL SOLUTION:

Looking at 15! (pronounced 15 factorial), we probably won’t spend the time on test day to calculate that it can be rewritten as 1,307,674,368,000. However I already knew that it would have to end in exactly three 0’s. Why? Because 0 is only formed by multiplying 2 by 5. Thus every zero will indicate the presence of both a 5 and 2. The three 0’s are the result of multiplying 2’s and 5’s in the prime factors of these numbers. Unsurprisingly, there will be a lot of 2’s in this product, but only a handful of 5’s.

How many 5’s exactly? Well 5 is a prime number, so it won’t be formed by the product of two different factors. It will only appear in numbers that have a five in them, or the numbers that end in 5 or 0. In 15! there are three numbers that will have a 5 in them: 5, 10 and 15. Thus we know that this number is the product of 53 as well as several other prime numbers.

Solving this problem does not require us to take the full prime factorization of 15*14*13*12*11*10*9*8*7*6*5*4*3*2*1, but you certainly could do it if you wanted to. The answer is 211*36*53*72*111*131. (Fun fact, this number will have exactly 4032 factors… collect them all!). The only number we are interested in is that 53, which means that we can raise 5 to the third power and still get a factor of 15 factorial. Thus, 54 will not qualify.

Answer choice B is correct.

chetan2u:

If this number was 25!, we would solve it as following:

25/5+25/5^2=5+1=6

Correct?

What is the greatest power that 5 can be raised to so that the resulting number is a factor of 15!?

(A) 2
(B) 3
(C) 4
(D) 5
(E) 6

15! = 15 ... * 10 * .. * 5 ... * 1
Three multiples of 5 are there, hence

Answer B.

15/5 = 3
3/5 = 0

Hence, Answer must be (B) 3

Hello,

So if I had to find out the greatest power that 3 can be raised to so that the resulting number was a factor of 122!
Is the following way correct:
122/3 = 40
122/9 = 13
122/27 = 4
122/81 = 1

So, 40+13+4+1 = 58, greatest power is 58.

Also can I use the same method to find the trailing zeros in 50! ?

15! = 15x 14 x 13 x 12 x 11 x 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1

………..15…………………………………..10…………………………..5……………………….

……….(5x3)………………………………(5x2)……………………….(5x1)………………….

Ans:3(B)

** Take the quotient only:
15/5 = 3
+ 3/5 = 0 or no quotitient here

Thus, the greatest power of 5 is: 3+0=3. FINAL ANSWER IS (B)

** Consider a different question: What is the greatest power that 25 (=5^2) can be raised to so that the resulting number is a factor of 15!?

Again, take the quotient only (as if we consider the single power of 5) :
15/5 = 3
+ 3/5 = 0 or no quotitient here

Thus, the greatest power of 25 is: (3+0)/2 = 1.5 = 1. Note that we always need to round down for this type of question.

Posted from my mobile device

This is a trailing zeros question:

15! = 15*14*13...*12*11

\(\frac{15}{5} = 3\)

Answer is B.

Let us see how many 5s are present in 15! .
5 being a prime number, is present at 5,10 and 15 (that is at 3 places)when 15! is expanded.
Hence greatest power that 5 can be raised is 3.
Alternately, you can also count the number of 5s by successive division and the result shall be 3 in the case.(B)

Hope this helps.