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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).

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

Bunuel wrote:

LAST DIGIT OF A POWER

Determining the last digit of (xyz)^n:

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?

Re: Math: Number Theory [#permalink]
14 Jan 2014, 00:56

Expert's post

mandrake15 wrote:

Bunuel wrote:

LAST DIGIT OF A POWER

Determining the last digit of (xyz)^n:

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?

Any integer with 6 as its units digit in any positive integer power has the units digit of 6 (integers ending with 0, 1, 5 or 6, in the integer power k>0, has the same last digit as the base.). For example, (xxx6)^(positive integer) has the units digit of 6.

The reason you get 0 as the units digit of (456)^35 is because it's a huge number and simple calculator rounds the result.

Exact result is: 1,158,162,485,059,181,044,784,824,077,056,791,483,879,723,809,565,243,305,114,019,731,744,476,935,058,125,438,332,149,170,176.