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Find the power of 80 in 40!??? [#permalink]
 18 Oct 2010, 18:17
This is regarding the topic "Factorial" written by Bunuel for GMAT Club's Math Book. When finding the power of non-prime number in n! we first do prime-factorization of the non-prime number and then find the powers of each prime number in n! one by one using the following formula \frac{n}{p}+\frac{{n}}{{p^2}}+\frac{{n}}{{p^3}}+....+\frac{{n}}{{p^x}} such that p^x <n, where p is the prime number. Let's suppose, we want to find the powers of 80 in 40!. Prime factorization of 80=2^4 * 5^1. Now first find the power of 2 in 40!; \frac{{40}}{{2}}+\frac{{40}}{{2^2}}+\frac{{40}}{{2^3}}+\frac{{40}}{{2^4}}+\frac{{40}}{{2^5}}=20+10+5+2+1=38powers of 2 in 40! --> 2^{38} Now find the powers of 5 in 40!; \frac{{40}}{{5}}+\frac{{40}}{{5^2}}=8+1=9 --> 5^9And 40!=80^x*q=(2^4 * 5^1)^x*q, where q is the quotient and x is any power of 80, now from above calculation 40!=(2^{38}*5^9)*q=(2^4*5^1)^9*2^2*q=(80)^9*4q, So we have 80 in the power of 9 in 40!. Now, the main reason for why did I do all of the above is that whether I am doing it right or not? I took values ( 80 & 40!) randomly and tried to apply the rule/method and I was confused whether I am (or should it be am I?  ) right or not. Would like expert opinions. -->Bunuel? or Shrouded1? or Gurpreetsingh?
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Re: Find the power of 80 in 40!??? [#permalink]
18 Oct 2010, 20:18
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Re: Find the power of 80 in 40!??? [#permalink]
18 Oct 2010, 21:55
shrouded1 wrote: Well done Perfect ! Posted from my mobile device  Thanks! man That's a relief
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Re: Find the power of 80 in 40!??? [#permalink]
18 Oct 2010, 23:12
wanted to ask whether questions like these have been asked
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Re: Find the power of 80 in 40!??? [#permalink]
19 Oct 2010, 13:42
AtifS wrote: This is regarding the topic "Factorial" written by Bunuel for GMAT Club's Math Book. When finding the power of non-prime number in n! we first do prime-factorization of the non-prime number and then find the powers of each prime number in n! one by one using the following formula \frac{n}{p}+\frac{{n}}{{p^2}}+\frac{{n}}{{p^3}}+....+\frac{{n}}{{p^x}} such that p^x <n, where p is the prime number. Let's suppose, we want to find the powers of 80 in 40!. Prime factorization of 80=2^4 * 5^1. Now first find the power of 2 in 40!; \frac{{40}}{{2}}+\frac{{40}}{{2^2}}+\frac{{40}}{{2^3}}+\frac{{40}}{{2^4}}+\frac{{40}}{{2^5}}=20+10+5+2+1=38powers of 2 in 40! --> 2^{38} Now find the powers of 5 in 40!; \frac{{40}}{{5}}+\frac{{40}}{{5^2}}=8+1=9 --> 5^9And 40!=80^x*q=(2^4 * 5^1)^x*q, where q is the quotient and x is any power of 80, now from above calculation 40!=(2^{38}*5^9)*q=(2^4*5^1)^9*2^2*q=(80)^9*4q, So we have 80 in the power of 9 in 40!. Now, the main reason for why did I do all of the above is that whether I am doing it right or not? I took values ( 80 & 40!) randomly and tried to apply the rule/method and I was confused whether I am (or should it be am I? ) right or not. Would like expert opinions. -->Bunuel? or Shrouded1? or Gurpreetsingh? Yes, that's correct. There is an example about power of 900 in 50! at: everything-about-factorials-on-the-gmat-85592-20.htmlmrinal2100 wrote: wanted to ask whether questions like these have been asked You'll need only to know how to determine the number of trailing zeros and the power of primes in n! ( everything-about-factorials-on-the-gmat-85592.html), the above example is out of the scope of GMAT.
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Re: Find the power of 80 in 40!??? [#permalink]
06 Sep 2011, 18:27
thanks guys..this example was very helpful
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Re: Find the power of 80 in 40!??? [#permalink]
13 Sep 2011, 04:27
nice explanation.
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Re: Find the power of 80 in 40!??? [#permalink]
25 Jun 2012, 00:11
AtifS wrote: This is regarding the topic "Factorial" written by Bunuel for GMAT Club's Math Book. When finding the power of non-prime number in n! we first do prime-factorization of the non-prime number and then find the powers of each prime number in n! one by one using the following formula \frac{n}{p}+\frac{{n}}{{p^2}}+\frac{{n}}{{p^3}}+....+\frac{{n}}{{p^x}} such that p^x <n, where p is the prime number. Let's suppose, we want to find the powers of 80 in 40!. Prime factorization of 80=2^4 * 5^1. Now first find the power of 2 in 40!; \frac{{40}}{{2}}+\frac{{40}}{{2^2}}+\frac{{40}}{{2^3}}+\frac{{40}}{{2^4}}+\frac{{40}}{{2^5}}=20+10+5+2+1=38powers of 2 in 40! --> 2^{38} Now find the powers of 5 in 40!; \frac{{40}}{{5}}+\frac{{40}}{{5^2}}=8+1=9 --> 5^9And 40!=80^x*q=(2^4 * 5^1)^x*q, where q is the quotient and x is any power of 80, now from above calculation
40!=(2^{38}*5^9)*q=(2^4*5^1)^9*2^2*q=(80)^9*4q, So we have 80 in the power of 9 in 40!. Now, the main reason for why did I do all of the above is that whether I am doing it right or not? I took values ( 80 & 40!) randomly and tried to apply the rule/method and I was confused whether I am (or should it be am I? ) right or not. Would like expert opinions. -->Bunuel? or Shrouded1? or Gurpreetsingh? Can u explain more the step in yellow, please? Thanks
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Re: Find the power of 80 in 40!??? [#permalink]
25 Jun 2012, 00:49
Hi Anan, Not to worry much as this particular concept has been explained very well was Bunuel in math book as well as in one of the topics related to factorial. You can find them here at everything-about-factorials-on-the-gmat-85592.html and math-number-theory-88376.html. If still concept remains unclear, please ask. AnanJammal wrote: AtifS wrote: This is regarding the topic "Factorial" written by Bunuel for GMAT Club's Math Book. When finding the power of non-prime number in n! we first do prime-factorization of the non-prime number and then find the powers of each prime number in n! one by one using the following formula \frac{n}{p}+\frac{{n}}{{p^2}}+\frac{{n}}{{p^3}}+....+\frac{{n}}{{p^x}} such that p^x <n, where p is the prime number. Let's suppose, we want to find the powers of 80 in 40!. Prime factorization of 80=2^4 * 5^1. Now first find the power of 2 in 40!; \frac{{40}}{{2}}+\frac{{40}}{{2^2}}+\frac{{40}}{{2^3}}+\frac{{40}}{{2^4}}+\frac{{40}}{{2^5}}=20+10+5+2+1=38powers of 2 in 40! --> 2^{38} Now find the powers of 5 in 40!; \frac{{40}}{{5}}+\frac{{40}}{{5^2}}=8+1=9 --> 5^9And 40!=80^x*q=(2^4 * 5^1)^x*q, where q is the quotient and x is any power of 80, now from above calculation
40!=(2^{38}*5^9)*q=(2^4*5^1)^9*2^2*q=(80)^9*4q, So we have 80 in the power of 9 in 40!. Now, the main reason for why did I do all of the above is that whether I am doing it right or not? I took values ( 80 & 40!) randomly and tried to apply the rule/method and I was confused whether I am (or should it be am I? ) right or not. Would like expert opinions. -->Bunuel? or Shrouded1? or Gurpreetsingh? Can u explain more the step in yellow, please? Thanks |
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Re: Find the power of 80 in 40!???
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25 Jun 2012, 00:49
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