sandeep800 wrote:

AndreG wrote:

What is the unit's digit of \(7^{75} + 6\) ?

(C) 2008 GMAT Club -

m12#29 * 1

* 3

* 5

* 7

* 9

I put the official explanation and the part I do not understand (blue text) in a spoiler

\(7^1\) ends with 7

\(7^2\) ends with 9

\(7^3\) ends with 3

\(7^4\) ends with 1

\(7^5\) ends with 7

...

\(7^{76}\) ends with 1. --> ???

So, \(7^{75}\) ends with 3. --> ???

If 7^5 ends with 7, shouldnt 7^75 also end with 7? Hence 7+6=13 - answer b?? Please help!!

\(7^{75} + 6\) ends with 9.

The correct answer is E.

well i will say that whatever may be the number if we have to find the last digit of some number whose power isgiven..then the best method is to divide the power by 4 since all the digits from 1...9 will surely repeat after every 4th digit...

then raise the digit to the power of remainder...here 75/4 remainder=3

7^3=last digit comes out to be 3

now 3+6=9

thanx

The above is correct with a little correction: when remainder is zero, then we should rise to the power not of remainder 0 but to the power of the cyclicity number.

For example las digit of 7^24 is the same as the last digit of 7^4 as the cyclicity of 7 in power is 4 and 24 divided by 4 gives remainder of zero.

From Number Theory chapter of Math Book:

LAST DIGIT OF A POWERDetermining 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 \(3\).

Hope it helps.

Thanx a lot bunuel for correcting me..i wud have applied my method in GMAT if u had not corrected me....