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  The terminal zeros of a number are the zeros to the right of its last  [#permalink]
  
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  09 Oct 2018, 18:34
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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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  "Only $149 for 3 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Director Joined: 18 Jul 2018 Posts: 679 Location: India Concentration: Finance, Marketing WE: Engineering (Energy and Utilities) The terminal zeros of a number are the zeros to the right of its last [#permalink] Show Tags Updated on: 09 Oct 2018, 18:52 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 Posted from my mobile device _________________ Press +1 Kudo If my post helps! Originally posted by Afc0892 on 09 Oct 2018, 18:43. Last edited by Afc0892 on 09 Oct 2018, 18:52, edited 1 time in total. Math Expert Joined: 02 Aug 2009 Posts: 7335 Re: The terminal zeros of a number are the zeros to the right of its last [#permalink] Show Tags 09 Oct 2018, 18:44 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 _________________ 1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372 2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html 3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html 4) Base while finding % increase and % decrease : https://gmatclub.com/forum/percentage-increase-decrease-what-should-be-the-denominator-287528.html GMAT Expert Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 6985 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: The terminal zeros of a number are the zeros to the right of its last [#permalink] Show Tags 11 Oct 2018, 00:34 => 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. _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$149 for 3 month Online Course"
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  Re: The terminal zeros of a number are the zeros to the right of its last  [#permalink]
  
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  11 Oct 2018, 06:37
  MathRevolution wrote:
  [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?
  
  $$1) n^2 – 15n + 50 < 0$$
  $$2) n > 5$$
  
  $$?\,\,\,:\,\,\,{\text{number}}\,\,{\text{of}}\,\,{\text{terminal}}\,\,{\text{zeros}}\,\,{\text{of}}\,\,n\,{\text{!}}$$
  
  $$\left( 1 \right)\,\,\,{n^2} - 15n + 50 < 0\,\,\,\, \Leftrightarrow \,\,\,\,5 < n < 10$$
  
  $$\left. \begin{gathered} n! = 6! = 6 \cdot \boxed5 \cdot 4 \cdot 3 \cdot 2 \cdot 1\,\,\,\,\, \Rightarrow \,\,\,\,\,\frac{{6!}}{{10}} = \operatorname{int} \,\,\,\,\,{\text{but}}\,\,\,\,\,\frac{{6!}}{{{{10}^2}}} \ne \operatorname{int} \,\,\,\,\,\, \Rightarrow \,\,\,\,\,? = 1\,\, \hfill \\ n! = 7! = 7 \cdot 6 \cdot \boxed5 \cdot 4 \cdot 3 \cdot 2 \cdot 1\,\,\,\,\, \Rightarrow \,\,\,\,\,\frac{{7!}}{{10}} = \operatorname{int} \,\,\,\,\,{\text{but}}\,\,\,\,\,\frac{{7!}}{{{{10}^2}}} \ne \operatorname{int} \,\,\,\,\,\, \Rightarrow \,\,\,\,\,? = 1\,\, \hfill \\ n! = 8! = 8 \cdot 7 \cdot 6 \cdot \boxed5 \cdot 4 \cdot 3 \cdot 2 \cdot 1\,\,\,\,\, \Rightarrow \,\,\,\,\,\frac{{8!}}{{10}} = \operatorname{int} \,\,\,\,\,{\text{but}}\,\,\,\,\,\frac{{8!}}{{{{10}^2}}} \ne \operatorname{int} \,\,\,\,\,\, \Rightarrow \,\,\,\,\,? = 1\, \hfill \\ n! = 9! = 9 \cdot 8 \cdot 7 \cdot 6 \cdot \boxed5 \cdot 4 \cdot 3 \cdot 2 \cdot 1\,\,\,\,\, \Rightarrow \,\,\,\,\,\frac{{9!}}{{10}} = \operatorname{int} \,\,\,\,\,{\text{but}}\,\,\,\,\,\frac{{9!}}{{{{10}^2}}} \ne \operatorname{int} \,\,\,\,\,\, \Rightarrow \,\,\,\,\,? = 1\,\, \hfill \\ \end{gathered} \right\}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,? = 1$$
  
  $$\left( 2 \right)\,\,n > 5\,\,\,\,\,\left\{ \begin{gathered} \,{\text{Take}}\,\,n = 6\,\,\,\, \Rightarrow \,\,\,? = 1\,\, \hfill \\ \,{\text{Take}}\,\,n = 10\,\, \Rightarrow \,\,{\text{?}}\,\,{\text{ = }}\,\,{\text{2}}\,\, \hfill \\ \end{gathered} \right.$$
  
  This solution follows the notations and rationale taught in the GMATH method.
  
  Regards,
  Fabio.
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  Re: The terminal zeros of a number are the zeros to the right of its last  [#permalink]
  
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  11 Oct 2018, 19:03
  chetan2u wrote:
  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
  
  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.
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  Re: The terminal zeros of a number are the zeros to the right of its last  [#permalink]
  
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  12 Oct 2018, 00:13
  AnkitOrYadav wrote:
  chetan2u wrote:
  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
  
  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.
  
  Have the same doubt. Experts,kindly enlighten. Thank you. Bunuel VeritasKarishma
  Director
  Joined: 18 Jul 2018
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  The terminal zeros of a number are the zeros to the right of its last  [#permalink]
  
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  12 Oct 2018, 00:19
  AnkitOrYadav wrote:
  chetan2u wrote:
  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
  
  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.
  
  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
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  Re: The terminal zeros of a number are the zeros to the right of its last  [#permalink]
  
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  12 Oct 2018, 00:33
  Afc0892 wrote:
  AnkitOrYadav wrote:
  chetan2u wrote:
  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
  
  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.
  
  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
  
  Afc0892 thank you for the explanation!
  Re: The terminal zeros of a number are the zeros to the right of its last   [#permalink] 12 Oct 2018, 00:33
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explanation: The provided text is a GMAT math question from Math Revolution. It describes a problem related to terminal zeros in factorials and provides two conditions to determine the number of terminal zeros in \(n!\).
  
  1. The first condition is a quadratic inequality: \(n^2 – 15n + 50 < 0\). Solving this inequality, we find that \(5 < n < 10\). For \(n = 6, 7, 8, 9\), the number of terminal zeros in \(n!\) is 1.
  2. The second condition is \(n > 5\). This condition alone is not sufficient because different values of \(n\) (e.g., \(n = 6\) and \(n = 10\)) yield different numbers of terminal zeros.
  
  The document contains a question and an answer to the question.
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  The terminal zeros of a number are the zeros to the right of its last  [#permalink]
  
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  09 Oct 2018, 18:34
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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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  "Only $149 for 3 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Director Joined: 18 Jul 2018 Posts: 679 Location: India Concentration: Finance, Marketing WE: Engineering (Energy and Utilities) The terminal zeros of a number are the zeros to the right of its last [#permalink] Show Tags Updated on: 09 Oct 2018, 18:52 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 Posted from my mobile device _________________ Press +1 Kudo If my post helps! Originally posted by Afc0892 on 09 Oct 2018, 18:43. Last edited by Afc0892 on 09 Oct 2018, 18:52, edited 1 time in total. Math Expert Joined: 02 Aug 2009 Posts: 7335 Re: The terminal zeros of a number are the zeros to the right of its last [#permalink] Show Tags 09 Oct 2018, 18:44 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 _________________ 1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372 2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html 3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html 4) Base while finding % increase and % decrease : https://gmatclub.com/forum/percentage-increase-decrease-what-should-be-the-denominator-287528.html GMAT Expert Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 6985 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: The terminal zeros of a number are the zeros to the right of its last [#permalink] Show Tags 11 Oct 2018, 00:34 => 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. _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$149 for 3 month Online Course"
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  Re: The terminal zeros of a number are the zeros to the right of its last  [#permalink]
  
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  11 Oct 2018, 06:37
  MathRevolution wrote:
  [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?
  
  $$1) n^2 – 15n + 50 < 0$$
  $$2) n > 5$$
  
  $$?\,\,\,:\,\,\,{\text{number}}\,\,{\text{of}}\,\,{\text{terminal}}\,\,{\text{zeros}}\,\,{\text{of}}\,\,n\,{\text{!}}$$
  
  $$\left( 1 \right)\,\,\,{n^2} - 15n + 50 < 0\,\,\,\, \Leftrightarrow \,\,\,\,5 < n < 10$$
  
  $$\left. \begin{gathered} n! = 6! = 6 \cdot \boxed5 \cdot 4 \cdot 3 \cdot 2 \cdot 1\,\,\,\,\, \Rightarrow \,\,\,\,\,\frac{{6!}}{{10}} = \operatorname{int} \,\,\,\,\,{\text{but}}\,\,\,\,\,\frac{{6!}}{{{{10}^2}}} \ne \operatorname{int} \,\,\,\,\,\, \Rightarrow \,\,\,\,\,? = 1\,\, \hfill \\ n! = 7! = 7 \cdot 6 \cdot \boxed5 \cdot 4 \cdot 3 \cdot 2 \cdot 1\,\,\,\,\, \Rightarrow \,\,\,\,\,\frac{{7!}}{{10}} = \operatorname{int} \,\,\,\,\,{\text{but}}\,\,\,\,\,\frac{{7!}}{{{{10}^2}}} \ne \operatorname{int} \,\,\,\,\,\, \Rightarrow \,\,\,\,\,? = 1\,\, \hfill \\ n! = 8! = 8 \cdot 7 \cdot 6 \cdot \boxed5 \cdot 4 \cdot 3 \cdot 2 \cdot 1\,\,\,\,\, \Rightarrow \,\,\,\,\,\frac{{8!}}{{10}} = \operatorname{int} \,\,\,\,\,{\text{but}}\,\,\,\,\,\frac{{8!}}{{{{10}^2}}} \ne \operatorname{int} \,\,\,\,\,\, \Rightarrow \,\,\,\,\,? = 1\, \hfill \\ n! = 9! = 9 \cdot 8 \cdot 7 \cdot 6 \cdot \boxed5 \cdot 4 \cdot 3 \cdot 2 \cdot 1\,\,\,\,\, \Rightarrow \,\,\,\,\,\frac{{9!}}{{10}} = \operatorname{int} \,\,\,\,\,{\text{but}}\,\,\,\,\,\frac{{9!}}{{{{10}^2}}} \ne \operatorname{int} \,\,\,\,\,\, \Rightarrow \,\,\,\,\,? = 1\,\, \hfill \\ \end{gathered} \right\}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,? = 1$$
  
  $$\left( 2 \right)\,\,n > 5\,\,\,\,\,\left\{ \begin{gathered} \,{\text{Take}}\,\,n = 6\,\,\,\, \Rightarrow \,\,\,? = 1\,\, \hfill \\ \,{\text{Take}}\,\,n = 10\,\, \Rightarrow \,\,{\text{?}}\,\,{\text{ = }}\,\,{\text{2}}\,\, \hfill \\ \end{gathered} \right.$$
  
  This solution follows the notations and rationale taught in the GMATH method.
  
  Regards,
  Fabio.
  _________________
  
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  Re: The terminal zeros of a number are the zeros to the right of its last  [#permalink]
  
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  11 Oct 2018, 19:03
  chetan2u wrote:
  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
  
  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.
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  Re: The terminal zeros of a number are the zeros to the right of its last  [#permalink]
  
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  12 Oct 2018, 00:13
  AnkitOrYadav wrote:
  chetan2u wrote:
  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
  
  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.
  
  Have the same doubt. Experts,kindly enlighten. Thank you. Bunuel VeritasKarishma
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  The terminal zeros of a number are the zeros to the right of its last  [#permalink]
  
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  12 Oct 2018, 00:19
  AnkitOrYadav wrote:
  chetan2u wrote:
  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
  
  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.
  
  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
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  Re: The terminal zeros of a number are the zeros to the right of its last  [#permalink]
  
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  12 Oct 2018, 00:33
  Afc0892 wrote:
  AnkitOrYadav wrote:
  chetan2u wrote:
  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
  
  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.
  
  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
  
  Afc0892 thank you for the explanation!
  Re: The terminal zeros of a number are the zeros to the right of its last   [#permalink] 12 Oct 2018, 00:33
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distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The provided text is a GMAT math question from Math Revolution. It describes a problem related to terminal zeros in factorials and provides two conditions to determine the number of terminal zeros in \(n!\).
  
  1. The first condition is a quadratic inequality: \(n^2 – 15n + 50 < 0\). Solving this inequality, we find that \(5 < n < 10\). For \(n = 6, 7, 8, 9\), the number of terminal zeros in \(n!\) is 1.
  2. The second condition is \(n > 5\). This condition alone is not sufficient because different values of \(n\) (e.g., \(n = 6\) and \(n = 10\)) yield different numbers of terminal zeros.
  
  The document contains a question and an answer to the question.
  
  Label: question-and-answer