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

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

Chethan92 · Answer

From statement 1: solving the quadratic equation gives (n-5)(n-10)<0.

Or 5<n<10.

Max value of n can be 9. And 9! has 1 terminating zero.
A is sufficient.

From statement 2: n>5. But n can take any value. If n is 9, then 9! Will have 1 zero. If n is 11, then 11! Will have 2 zeros.
Hence B is insufficient.

A is the answer

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chetan2u · Answer

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 
 n^2-10n-5n+50<0......(n-10)(n-5)<0 
So n is between 5 and 10
For each of 6!,7!,8! or 9!, Number of zeroes will be 1 as only 1*5 is there in each
Sufficient

2) n > 5 
If 6 ANS is 1 and if 10, Ans is 2
Insufficient

A

MathRevolution · Answer

=>

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.

fskilnik · Answer

?\,\,\,:\,\,\,{	ext{number}}\,\,{	ext{of}}\,\,{	ext{terminal}}\,\,{	ext{zeros}}\,\,{	ext{of}}\,\,n\,{	ext{!}}

\left( 1 ight)\,\,\,{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} \,\,\,\,\,{	ext{but}}\,\,\,\,\,\frac{{6!}}{{{{10}^2}}} 
e \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} \,\,\,\,\,{	ext{but}}\,\,\,\,\,\frac{{7!}}{{{{10}^2}}} 
e \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} \,\,\,\,\,{	ext{but}}\,\,\,\,\,\frac{{8!}}{{{{10}^2}}} 
e \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} \,\,\,\,\,{	ext{but}}\,\,\,\,\,\frac{{9!}}{{{{10}^2}}} 
e \operatorname{int} \,\,\,\,\,\, \Rightarrow \,\,\,\,\,? = 1\,\, \hfill \ 
\end{gathered} ight\}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,? = 1

\left( 2 ight)\,\,n > 5\,\,\,\,\,\left\{ \begin{gathered}
 \,{	ext{Take}}\,\,n = 6\,\,\,\, \Rightarrow \,\,\,? = 1\,\, \hfill \
 \,{	ext{Take}}\,\,n = 10\,\, \Rightarrow \,\,{	ext{?}}\,\,{	ext{ = }}\,\,{	ext{2}}\,\, \hfill \ 
\end{gathered} ight.

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

Regards, 
Fabio.

StudiosTom · Answer

n2−10n−5n+50<0......(n−10)(n−5)<0
How did we solve this to get 5<n<10. Wont it be n<5 and n<10.Please help me solve these type of inequality.

poojasingavi · Answer

Have the same doubt. Experts,kindly enlighten. Thank you.  Bunuel   VeritasKarishma

Chethan92 · Answer

The highlighted portion is an inequality. solving gives n = 5 and 10. plotting this on the number line using wavy curve method. Refer to the attached file.

Since the inequality sign is <0. we should consider the integers in the range 5<n<10.

Refer this topic  https://gmatclub.com/forum/wavy-line-method-application-complex-algebraic-inequalities-224319.html  for wavy curve method

poojasingavi · Answer

Afc0892 thank you for the explanation!

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

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