Max@Math Revolution wrote:[Math Revolution GMAT math practice question]

If n is the greatest positive integer for which 5^n is a factor of 50!, what is the value of n?

A. 10
B. 11
C. 12
D. 13
E. 14

\[n \geqslant \,\,1\,\,\,\operatorname{int} \]
\[\frac{{50!}}{{{5^n}}} = \operatorname{int} \]
\[?\,\, = \,\,\max \,\,n\,\]

In English: how many primes equal to 5 are we able to find in the product 50*49*48*47*...*6*5*4*3*2*1 ?

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In 5 = 5*1 we find the first
In 10 = 5*2 we find the second
In 15 = 5*3 we find the third
In 20 = 5*4 we find the four
In 25 = 5*5 we find the fifth and the sixth, but forget the sixth for a moment, please!
In 30 = 5*6 we find the "sixth" (yes, that´s a lie... wait a bit!)
In 35 = 5*7 we find the "seventh" (wait...)
--- etc ---
In 50 = 5*10 we find the "tenth" (second mistake... because 5*10 = 5*5*2 .... but wait...)
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What is going on here?

When we calculate 50/5 = 10 , we find the "first" 5´s involved... with some lies...

When we divide 50 by 5^2 , we find exactly the numbers like 25 (and 50) , in which there are (at least) two 5´s in it, and only one 5 (in each case) was counted previously (between the parallel lines) ...
Now they were properly counted... no more lies!

Conclusion: 50/5 + 50/25 = 12 is the right answer!

(In this case, there are no 5^3 , 5^4 , ... in 50! . In other words, each of the integers 1, 2, 3, 4, 5, ... , 50 has at most two 5´s in its corresponding prime decomposition!)

Now the "recipe":
\[? = \left\lfloor {\frac{{50}}{5}} \right\rfloor + \left\lfloor {\frac{{50}}{{{5^2}}}} \right\rfloor + \left\lfloor {\frac{{50}}{{{5^3}}}} \right\rfloor + \ldots = 10 + 2 + 0 + 0 + 0 + \ldots = \boxed{12}\]

where \[\left\lfloor N \right\rfloor \] denotes the "floor" of N, that is, the greatest integer that is less than, or equal to, N.
If you prefer: when we divide 50 by 5, we have quotient 10 (the floor) and the remainder 0 (irrelevant when looking for the floor)!

Another example:

GMATH wrote: If n is the greatest positive integer for which 3^n is a factor of 50!, what is the value of n?

A. 19
B. 20
C. 21
D. 22
E. 23

Answer:
\[? = \left\lfloor {\frac{{50}}{3}} \right\rfloor + \left\lfloor {\frac{{50}}{{{3^2}}}} \right\rfloor + \left\lfloor {\frac{{50}}{{{3^3}}}} \right\rfloor + \left\lfloor {\frac{{50}}{{{3^4}}}} \right\rfloor + \ldots = 16 + 5 + 1 + 0 + \ldots = \boxed{22}\]

Fabio, do you think it is useful to remember this "recipe" for GMAT purposes?
YES, although it´s obviously unprobable you will need it.
(This is MUCH less important than to know that the [length of the] height of the equilateral triangle is the [length of the] side times half the square root of 3, for instance.)

But... If you understood the recipe, it will be much easier to remember it (and to apply it) correctly...

That´s the GMATH´s method "essence": REAL AND DEEP UNDERSTANDING!

Regards,
fskilnik.

=>
50! = 1*2*...*5*...*10*...*15*...*20*...*25*...*30*...*35*...*40*...*45*...*50
=1*2*...*(5)...*(2*5)*...*(3*5)*...*(4*5)*...*(52)*...*(6*5)*...*(7*5)*...*(8*5)*...*(9*5)*...*(2*5^2)
Since 5 is a prime number, no further factors of 5 appear in the prime factorization of 50!. The number of 5s in the above expansion of 50! is 12.

Therefore, the answer is C.
Answer: C

Hi,

This is a question which tests the maximum power concept in a factorial.

i.e., Counting the number of 5's in a factorial.

50! = 50 * 49 * 48 * 47 * 46 * 45 * ......3 * 2 *1.

To count 5's we can see that multiples of 5 is what we need to focus, but we need to be careful some multiples of 5's have more than one 5. Like in this case, 25 - 5^2 and 50 - 5^2 * 2

To count the number of five's in a 50 factorial,

We need to count in the number of 5's and number of 5^2's

50/ 5 + 50 / 5^2 = 10 + 2 = 12

So the maximum power of n has to be 12. So the answer is C.

Above method is applicable for any prime number. If the number is not a prime number then we have to do the prime factorization and then find the result according to the question.

Hope it helps.