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Re: Math: Number Theory [#permalink]
04 Sep 2012, 18:34
Bunuel wrote:
NUMBER THEORY
Trailing zeros: Trailing zeros are a sequence of 0's in the decimal representation (or more generally, in any positional representation) of a number, after which no other digits follow.
125000 has 3 trailing zeros;
The number of trailing zeros in the decimal representation of n!, the factorial of a non-negative integer \(n\), can be determined with this formula:
\(\frac{n}{5}+\frac{n}{5^2}+\frac{n}{5^3}+...+\frac{n}{5^k}\), where k must be chosen such that \(5^k<n\).
It's easier if you look at an example:
How many zeros are in the end (after which no other digits follow) of \(32!\)? \(\frac{32}{5}+\frac{32}{5^2}=6+1=7\) (denominator must be less than 32, \(5^2=25\) is less)
Hence, there are 7 zeros in the end of 32!
The formula actually counts the number of factors 5 in n!, but since there are at least as many factors 2, this is equivalent to the number of factors 10, each of which gives one more trailing zero.
I noticed in case the number (n) is multiple of \(5^k\) and we have to find number of trailing zero zeroes, then it will be \(5^k<=n\) rather \(5^k<n\)
Re: Math: Number Theory [#permalink]
04 Sep 2012, 23:13
conty911 wrote:
Bunuel wrote:
NUMBER THEORY
Trailing zeros: Trailing zeros are a sequence of 0's in the decimal representation (or more generally, in any positional representation) of a number, after which no other digits follow.
125000 has 3 trailing zeros;
The number of trailing zeros in the decimal representation of n!, the factorial of a non-negative integer \(n\), can be determined with this formula:
\(\frac{n}{5}+\frac{n}{5^2}+\frac{n}{5^3}+...+\frac{n}{5^k}\), where k must be chosen such that \(5^k<n\).
It's easier if you look at an example:
How many zeros are in the end (after which no other digits follow) of \(32!\)? \(\frac{32}{5}+\frac{32}{5^2}=6+1=7\) (denominator must be less than 32, \(5^2=25\) is less)
Hence, there are 7 zeros in the end of 32!
The formula actually counts the number of factors 5 in n!, but since there are at least as many factors 2, this is equivalent to the number of factors 10, each of which gives one more trailing zero.
I noticed in case the number (n) is multiple of \(5^k\) and we have to find number of trailing zero zeroes, then it will be \(5^k<=n\) rather \(5^k<n\)
no of trailing zeros in 25! =6
\(\frac{25}{5}+\frac{25}{5^2}= 5+1\); Please correct me, clarify if i'm wrong. Thanks
You are right. \(k\) is the highest power of 5 not exceeding \(n.\) _________________
PhD in Applied Mathematics Love GMAT Quant questions and running.
Re: Math: Number Theory [#permalink]
12 Dec 2012, 02:26
Hi,
I'm not sure whether I undertood the below rule correctly:
"Integer ending with 0, 1, 5 or 6, in the integer power k>0, has the same last digit as the base".
55^2 = 3025 - the last digit is same as the base (5) so the above rule works. 55^10 = 253295162119141000 - the last digit is not same as the base (5) so the above rule doesn't work.
Re: Math: Number Theory [#permalink]
12 Dec 2012, 02:33
Expert's post
klueless7825 wrote:
Hi,
I'm not sure whether I undertood the below rule correctly:
"Integer ending with 0, 1, 5 or 6, in the integer power k>0, has the same last digit as the base".
55^2 = 3025 - the last digit is same as the base (5) so the above rule works. 55^10 = 25329516211914100[b]0[/b] - the last digit is not same as the base (5) so the above rule doesn't work.
Please help if I have misunderstood the rule.
5 in any positive integer power has 5 as the units digit.
5^1=5; 5^2=25; 5^3=125 ... 5^10=253,295,162,119,140,625 (your result was just rounded).
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