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HELP

by AJWILL » Fri Jul 13, 2012 2:18 am
If n is a positive integer greater than 2, n=?

1)the ten's digit of 11^2 = 4
2)the hundred's of 5^n = 6

please suggest a technical way to do this. i do not want to attempt this by putting values.

thanks
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by Anurag@Gurome » Fri Jul 13, 2012 2:44 am
AJWILL wrote:If n is a positive integer greater than 2, n=?

1)the ten's digit of 11^2 = 4
2)the hundred's of 5^n = 6
I assume the red part will be 11^n.
Otherwise that statement doesn't make any sense.

Statement 1: The ten's digit of 11^n will always be the unit's digit of n.
Hence, unit's digit of n is 4.

Not sufficient

Statement 2: Hundred's digit of 5^n will be 6 only when n is an even integer greater than 2.

Not sufficient.

1 & 2 Together: n can be 4, 14, 24 etc.

Not sufficient

The correct answer is E.

Note : For proofs of the underlined statements see my post below.
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by AJWILL » Fri Jul 13, 2012 2:55 am
thank you for these number properties but where can i check for proofs?
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by Anurag@Gurome » Fri Jul 13, 2012 3:01 am
The ten's digit of 11^n will always be the unit's digit of n
We can see the pattern as follows,
  • 11^1 = 11 ......... Ten's digit = 1
    11^2 = 121 ........ Ten's digit = 2
    11^3 = 1331 ....... Ten's digit = 3
    11^4 = 14641 ...... Ten's digit = 4
    ...
Or we can use binomial expansion as follows...
11^n = (10 + 1)^n = (Terms multiple of 10^2 + (nC1)*(10^1)*(1^(n - 1)) + 1)
Hence, ten's digit of 11^n will depend upon (nC1)*(10^1)*(1^(n - 1)) + 1 = 10(nC1) + 1 = 10n + 1

Hence, 11^n = (Some number multiple of 100 + 10n + 1) = (100K + 10n + 1)

Hence, ten's digit of 11^n will always be n.


Hundred's digit of 5^n will be 6 only when n is an even integer greater than 2
5^3 = 125
5^4 = 625
5^5 = 3125

Now, note that last three digits 5*(number ending with 125) is 625 and last three digits of 5*(number ending with 625) is 125. Hence, for powers greater than 2, the last three digits of 5^n will be 125 for odd powers and 625 for even powers.
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