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The terminal zeros of a number are the zeros to the right of

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The terminal zeros of a number are the zeros to the right of

by Max@Math Revolution » Tue Oct 09, 2018 6:28 pm

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[Math Revolution GMAT math practice question]

The terminal zeros of a number are the zeros to the right of its last nonzero digit. For example, 30,500 has two terminal zeros because there are two zeros to the right of its last nonzero digit, 5. How many terminal zeros does n! have?

1) n^2 - 15n + 50 < 0
2) n > 5
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by fskilnik@GMATH » Wed Oct 10, 2018 5:03 am

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

The terminal zeros of an integer are the zeros to the right of its last nonzero digit. For example, 30,500 has two terminal zeros because there are two zeros to the right of its last nonzero digit, 5. How many terminal zeros does n! have?

1) n^2 - 15n + 50 < 0
2) n > 5
\[?\,\,\,:\,\,\,{\text{number}}\,\,{\text{of}}\,\,{\text{terminal}}\,\,{\text{zeros}}\,\,{\text{of}}\,\,n\,{\text{!}}\]
\[\left( 1 \right)\,\,\,{n^2} - 15n + 50 < 0\,\,\,\, \Leftrightarrow \,\,\,\,5 < n < 10\]
\[\left. \begin{gathered}
n! = 6! = 6 \cdot \boxed5 \cdot 4 \cdot 3 \cdot 2 \cdot 1\,\,\,\,\, \Rightarrow \,\,\,\,\,\frac{{6!}}{{10}} = \operatorname{int} \,\,\,\,\,{\text{but}}\,\,\,\,\,\frac{{6!}}{{{{10}^2}}} \ne \operatorname{int} \,\,\,\,\,\, \Rightarrow \,\,\,\,\,? = 1\,\, \hfill \\
n! = 7! = 7 \cdot 6 \cdot \boxed5 \cdot 4 \cdot 3 \cdot 2 \cdot 1\,\,\,\,\, \Rightarrow \,\,\,\,\,\frac{{7!}}{{10}} = \operatorname{int} \,\,\,\,\,{\text{but}}\,\,\,\,\,\frac{{7!}}{{{{10}^2}}} \ne \operatorname{int} \,\,\,\,\,\, \Rightarrow \,\,\,\,\,? = 1\,\, \hfill \\
n! = 8! = 8 \cdot 7 \cdot 6 \cdot \boxed5 \cdot 4 \cdot 3 \cdot 2 \cdot 1\,\,\,\,\, \Rightarrow \,\,\,\,\,\frac{{8!}}{{10}} = \operatorname{int} \,\,\,\,\,{\text{but}}\,\,\,\,\,\frac{{8!}}{{{{10}^2}}} \ne \operatorname{int} \,\,\,\,\,\, \Rightarrow \,\,\,\,\,? = 1\, \hfill \\
n! = 9! = 9 \cdot 8 \cdot 7 \cdot 6 \cdot \boxed5 \cdot 4 \cdot 3 \cdot 2 \cdot 1\,\,\,\,\, \Rightarrow \,\,\,\,\,\frac{{9!}}{{10}} = \operatorname{int} \,\,\,\,\,{\text{but}}\,\,\,\,\,\frac{{9!}}{{{{10}^2}}} \ne \operatorname{int} \,\,\,\,\,\, \Rightarrow \,\,\,\,\,? = 1\,\, \hfill \\
\end{gathered} \right\}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,? = 1\]
\[\left( 2 \right)\,\,n > 5\,\,\,\,\,\left\{ \begin{gathered}
\,{\text{Take}}\,\,n = 6\,\,\,\, \Rightarrow \,\,\,? = 1\,\, \hfill \\
\,{\text{Take}}\,\,n = 10\,\, \Rightarrow \,\,{\text{?}}\,\,{\text{ = }}\,\,{\text{2}}\,\, \hfill \\
\end{gathered} \right.\]

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
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by Max@Math Revolution » Thu Oct 11, 2018 12:26 am

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=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Since we have 1 variable (n) and 0 equations, D is most likely to be the answer. So, we should consider each of the conditions on their own first.

Consider 1)
n^2 - 15n + 60 < 0
=> (n-5)(n-10) < 0
=> 5 < n < 10

The number of terminal zeros of a number is determined by the number of 5s in its prime factorization.

The integers satisfying 5 < n < 10 are 6, 7, 8 and 9. We count the 5s in the prime factorizations of 6!, 7!, 8! and 9!:
6! has one 5 in its prime factorization.
7! has one 5 in its prime factorization.
8! has one 5 in its prime factorization.
9! has one 5 in its prime factorization.

Thus, for 5 < n < 10, n! has one terminal zero.
As it gives us a unique answer, condition 1) is sufficient.

Condition 2)

If n = 6, then 6 > 5 and 6! = 720 has one terminal 0.
If n = 10, then 10 > 5 and 10! = 3,628,800 has two terminal 0s.

Condition 2) is not sufficient since it does not give a unique solution.

Therefore, A is the answer.
Answer: A

If the original condition includes "1 variable", or "2 variables and 1 equation", or "3 variables and 2 equations" etc., one more equation is required to answer the question. If each of conditions 1) and 2) provide an additional equation, there is a 59% chance that D is the answer, a 38% chance that A or B is the answer, and a 3% chance that the answer is C or E. Thus, answer D (conditions 1) and 2), when applied separately, are sufficient to answer the question) is most likely, but there may be cases where the answer is A,B,C or E.
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