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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 nonprime number in \(n!\) we first do primefactorization of the nonprime 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 nonprime number in \(n!\) we first do primefactorization of the nonprime 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: everythingaboutfactorialsonthegmat8559220.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! ( everythingaboutfactorialsonthegmat85592.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 nonprime number in \(n!\) we first do primefactorization of the nonprime 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 everythingaboutfactorialsonthegmat85592.html and mathnumbertheory88376.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 nonprime number in \(n!\) we first do primefactorization of the nonprime 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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Find the power of 80 in 40!??? [#permalink]
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14 Jul 2017, 17:32
Not sure if this is helpful to anyway, but this is how I think about these types of problems:
What are the components of 80? > factor 80 > (2^4)*(5).
So we are asking how many "lots" of four 2's and one 5 are there in 40? Well, there are a ton of 2's but 5's are the limiting factor. To phrase it a different way, I imagine a house that requires four 2's and one 5 for each house. Given that 5 is the limiting factor we can search for that.
40 > one 5 35 > one 5 30 > one 5 25 > two 5's 20 > one 5 15 > one 5 10 > one 5 5 > one 5
Total 5's = 9, therefore the total number of houses you can build is 9, therefore you can have nine 80's.



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Re: Find the power of 80 in 40!??? [#permalink]
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15 Jul 2017, 00:09
AtifS wrote: This is regarding the topic "Factorial" written by Bunuel for GMAT Club's Math Book. When finding the power of nonprime number in \(n!\) we first do primefactorization of the nonprime 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? We can also determine this by factoring the power required and then we can just find the power of the highest prime and dividing the result so obtained by the number of power of the highest prime so obtained by factoring the required number. Eg., In the above example, by factoring 80 = 2^4 * 5^1 Now, find the power of highest prime, i.e., 5 in 40! {(40/5) + (40/5^2)} {8 + (40/25)} {8 + 1} {9} Now, for 80, we need four 2's and one 5. Hence, the power of 80 in 40! = 9/1 (1 in denominator comes from number of 5's, which is the highest prime after factoring 80, required). Answer = 9



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Re: Find the power of 80 in 40!??? [#permalink]
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22 Aug 2017, 20:08
AtifS wrote: This is regarding the topic "Factorial" written by Bunuel for GMAT Club's Math Book. When finding the power of nonprime number in \(n!\) we first do primefactorization of the nonprime 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? hi here 5 is a limiting factor...so we can only work on 5, 5^9 .... thanks




Re: Find the power of 80 in 40!???
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