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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 \(5^k<n\)

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.

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.

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).

Awesome post. Bunuel, you are great. Love your posts.
_________________

GMAT RC Vocab - No nonsense(Only for GMAT) http://gmatclub.com/forum/gmat-rc-vocab-no-nonsense-only-for-gmat-162129.html#p1283165

Quant Document to revise a week before exam - Mixedbag http://gmatclub.com/forum/document-to-revise-a-week-before-exam-mixedbag-162145.html

Best questions to revise few days before exam- Mixed bag(25) http://gmatclub.com/forum/best-questions-to-revise-few-days-before-exam-mixed-bag-162124.html#p1283141

1. Last digit of \((xyz)^n\) is the same as that of \(z^n\); 2. Determine the cyclicity number \(c\) of \(z\); 3. Find the remainder \(r\) when \(n\) divided by the cyclisity; 4. When \(r>0\), then last digit of \((xyz)^n\) is the same as that of \(z^r\) and when \(r=0\), then last digit of \((xyz)^n\) is the same as that of \(z^c\), where \(c\) is the cyclisity number.

• Integer ending with 0, 1, 5 or 6, in the integer power k>0, has the same last digit as the base. • Integers ending with 2, 3, 7 and 8 have a cyclicity of 4. • Integers ending with 4 (eg. \((xy4)^n\)) have a cyclisity of 2. When n is odd \((xy4)^n\) will end with 4 and when n is even \((xy4)^n\) will end with 6. • Integers ending with 9 (eg. \((xy9)^n\)) have a cyclisity of 2. When n is odd \((xy9)^n\) will end with 9 and when n is even \((xy9)^n\) will end with 1.

Example: What is the last digit of \(127^{39}\)? Solution: Last digit of \(127^{39}\) is the same as that of \(7^{39}\). Now we should determine the cyclisity of \(7\):

1. 7^1=7 (last digit is 7) 2. 7^2=9 (last digit is 9) 3. 7^3=3 (last digit is 3) 4. 7^4=1 (last digit is 1) 5. 7^5=7 (last digit is 7 again!) ...

So, the cyclisity of 7 is 4.

Now divide 39 (power) by 4 (cyclisity), remainder is 3.So, the last digit of \(127^{39}\) is the same as that of the last digit of \(7^{39}\), is the same as that of the last digit of \(7^3\), which is \(3\).

Congratulation and thank you very much for the post, but in the LAST DIGIT OF A POWER i have an issue, when i try to solve the last digit of (456)^35 with the process i just don't get the correct answers, with the process above gives me 6^4 which is 1296=6 and with calculator its 0, can you explain me that case?

gmatclubot

Re: Math: Number Theory
[#permalink]
13 Jan 2014, 19:57

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