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Manager  Status: Not afraid of failures, disappointments, and falls.
Joined: 20 Jan 2010
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Concentration: Technology, Entrepreneurship
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Find the power of 80 in 40!???  [#permalink]

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4
11
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?
Retired Moderator Joined: 02 Sep 2010
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Re: Find the power of 80 in 40!???  [#permalink]

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Well done
Perfect !

Posted from my mobile device
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Manager  Status: Not afraid of failures, disappointments, and falls.
Joined: 20 Jan 2010
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Re: Find the power of 80 in 40!???  [#permalink]

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shrouded1 wrote:
Well done
Perfect !

Posted from my mobile device

Thanks! man
That's a relief Manager  Joined: 29 Sep 2008
Posts: 82
Re: Find the power of 80 in 40!???  [#permalink]

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Math Expert V
Joined: 02 Sep 2009
Posts: 60480
Re: Find the power of 80 in 40!???  [#permalink]

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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.html

mrinal2100 wrote:

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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thanks guys..this example was very helpful
Manager  Joined: 01 Nov 2010
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GMAT Date: 08-27-2012
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Re: Find the power of 80 in 40!???  [#permalink]

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nice explanation.
Intern  Joined: 24 Jun 2012
Posts: 4
GMAT Date: 08-08-2012
GPA: 3
Re: Find the power of 80 in 40!???  [#permalink]

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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
Manager  Joined: 07 Sep 2011
Posts: 59
Location: United States
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WE: General Management (Real Estate)
Re: Find the power of 80 in 40!???  [#permalink]

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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.

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
Intern  Joined: 01 Nov 2014
Posts: 1
Re: Find the power of 80 in 40!???  [#permalink]

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the step in yellow is unclear for me also Manager  Joined: 20 Jul 2013
Posts: 51
Re: Find the power of 80 in 40!???  [#permalink]

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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 * int

2^4 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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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!.

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Joined: 22 Jan 2017
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Find the power of 80 in 40!???  [#permalink]

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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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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?

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).

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Re: Find the power of 80 in 40!???  [#permalink]

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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?

hi

here 5 is a limiting factor...so we can only work on 5, 5^9 ....

thanks
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Re: Find the power of 80 in 40!???  [#permalink]

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i kept thinking........how do we know that 80 IS a factor of 40!? Re: Find the power of 80 in 40!???   [#permalink] 24 Nov 2019, 05:14
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