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Find the power of 80 in 40!??? [#permalink]
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 18 Oct 2010, 18:17
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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 \(40!\); \(\frac{{40}}{{2}}+\frac{{40}}{{2^2}}+\frac{{40}}{{2^3}}+\frac{{40}}{{2^4}}+\frac{{40}}{{2^5}}=20+10+5+2+1=38\)powers of \(2\) in \(40!\) --> \(2^{38}\) Now find the powers of \(5\) in \(40!\); \(\frac{{40}}{{5}}+\frac{{40}}{{5^2}}=8+1=9 --> 5^9\) And \(40!=80^x*q=(2^4 * 5^1)^x*q\), 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]
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18 Oct 2010, 20:18
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Re: Find the power of 80 in 40!??? [#permalink]
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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]
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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]
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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 \(40!\); \(\frac{{40}}{{2}}+\frac{{40}}{{2^2}}+\frac{{40}}{{2^3}}+\frac{{40}}{{2^4}}+\frac{{40}}{{2^5}}=20+10+5+2+1=38\)powers of \(2\) in \(40!\) --> \(2^{38}\) Now find the powers of \(5\) in \(40!\); \(\frac{{40}}{{5}}+\frac{{40}}{{5^2}}=8+1=9 --> 5^9\) And \(40!=80^x*q=(2^4 * 5^1)^x*q\), 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]
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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]
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13 Sep 2011, 04:27
nice explanation.
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Re: Find the power of 80 in 40!??? [#permalink]
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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 \(40!\); \(\frac{{40}}{{2}}+\frac{{40}}{{2^2}}+\frac{{40}}{{2^3}}+\frac{{40}}{{2^4}}+\frac{{40}}{{2^5}}=20+10+5+2+1=38\)powers of \(2\) in \(40!\) --> \(2^{38}\) Now find the powers of \(5\) in \(40!\); \(\frac{{40}}{{5}}+\frac{{40}}{{5^2}}=8+1=9 --> 5^9\) And \(40!=80^x*q=(2^4 * 5^1)^x*q\), 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]
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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 \(40!\); \(\frac{{40}}{{2}}+\frac{{40}}{{2^2}}+\frac{{40}}{{2^3}}+\frac{{40}}{{2^4}}+\frac{{40}}{{2^5}}=20+10+5+2+1=38\)powers of \(2\) in \(40!\) --> \(2^{38}\) Now find the powers of \(5\) in \(40!\); \(\frac{{40}}{{5}}+\frac{{40}}{{5^2}}=8+1=9 --> 5^9\) And \(40!=80^x*q=(2^4 * 5^1)^x*q\), 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]
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07 Dec 2014, 01:49
the step in yellow is unclear for me also 
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Re: Find the power of 80 in 40!??? [#permalink]
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31 Jan 2015, 04:27
nadooz wrote: the step in yellow is unclear for me also The key thing to see is that 40! / 80^n = Int. After factoring and finding the respective powers for 5 and 2 ... manipulate the original equation in the question stem to 40! = 80^n * int.So, we know that 40! will equal 80^n * some int. value (in this case, we'll mark that int. as "p" -- for simplicity's sake) Since 80^n = (5^9 * 2^38) --> We now have 40! = (5^9 * 2^38) * p Now, we need to match the higher power down to the lower power. We need to deduce 2^38 to the 9th power, so that it'll match up with 5^9. Knowing that when we factorized 80^n ... we were left with 5^n and 2^4n ... Using 2^4 --> we want to reduce the 38 down by dividing 4 into 38. Well, that obviously won't work. So, we take out 2^2: 40! = (5^9 * 2^36) * 2^2 * p Then, since we have 2^4 as a factor of 80, we plug that in for 2^36. We divide 4 into 36, and left with this ---> 40! = (5^9 * (2^4)^9) * 2^2 * int2^4 becomes 8: 40! = (5^9 * 16^9) * 2^2 * int -----> 40! = (5 * 16)^9 * 2^2 * int ----> 40! = (80)^9 * 2^2 * int So: n = 9
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Re: Find the power of 80 in 40!??? [#permalink]
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16 Feb 2015, 23:14
Since 80 = 16*5 = 2^4*5 (prime factorization)
40/2= (20) - 20/2=(10) - 10/2=(5) - 5/2=(2) - 2/2=(1) (only quotient without remainder)
Total powers of 2 in 40! = 20+10+5+2+1 = 38
So power of 2^4 in 4! = 38/4 = 9
Powers of 5 in 40! = 40/5=(8) - 8/5=(1)
8+1 = 9
Both 16 and 5 are having power of 9 in 40!.
Answer is 9.
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Re: Find the power of 80 in 40!??? [#permalink]
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Re: Find the power of 80 in 40!???
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