vitaliyGMAT wrote:
Good question.
\(N = 10^{40} + 2^{40}\) = \(2^{40}(5^{40} + 1)\)
\(5^{any-power}\) will always be odd, hence \(5^{40} + 1\) will be even.
The question is: can we get multiple of 4? Last digit of \(5^{40} + 1\) will be 6. When we divide that huge number \(...........6\) by \(2\) we'll get last digit \(3\) (odd) and the answer to the question is no.
Hence \(max\) power of \(2\) in \(N\) is \(41\).
\(k=41\), ----> \(k - 2 = 39\).
Answer B.
That's a solid solution. However, I'll point out one incorrect statement, in case it helps out with a similar question in the future.
You're correct to say that the last digit of 5^40 + 1 will be 6.
However, knowing that the units digit of a number is 6, does
not necessarily mean that, when we divide that number by 2, the units digit of the quotient will be 3.
For example, 1056 divided by 2 equals 528
Likewise, 96 divided by 2 equals 48
So, to remove any uncertainty, we must recognize that 5^b will end in 25 for all integer values of b greater than 1.
For example, 5^2 =
255^3 = 1
255^4 = 6
255^5 = 31
25 etc....
So, 5^40 + 1 = ?????
25 + 1 = ?????26
And when we divide ?????26 by 2, the quotient is ??????
3Cheers,
Brent
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