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Bunuel
If n is a positive integer such that n and 2^n have the same units digit, what is the sum of the smallest and second smallest possible values of n?

A. 6
B. 12
C. 24
D. 30
E. 48


It the cyclicity of the number 2 is written down, all we need to do
is to check which columns have the same units digit and add the smallest 2 numbers.

n-----1-----2-----3----4-----5-----6-----7-----8-----9----10----11----12---13---14----15----16
2^n--2-----4-----8----6-----2-----4-----8-----6-----2-----4-----8------6-----2----4-----8------6

Therefore, the sum of the smallest two values is \(14+26 = 30\)(Option D)
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Bunuel
If n is a positive integer such that n and 2^n have the same units digit, what is the sum of the smallest and second smallest possible values of n?

A. 6
B. 12
C. 24
D. 30
E. 48

It would be great to make a table here and we can see a nice pattern on \(2^n\). Note we only need to input even n's since the units digit of \(2^n\) has to be even.

n \(2^n\)
2 4
4 6
6 4
8 6
10 4
12 6
14 4
16 6


So the answer is 14 + 16 = 30.

Ans: D
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How n should be an even integer?. Is 2^n=n here?
niks18


\(2\) raised to any positive integer will be an even number so \(n\) has to be an even number.

Few even powers of \(2\) are
\(2^2=4\)
\(2^4=16\)
\(2^6=64\)
\(2^8=256\)

So we see that even powers of two has unit's digits either \(4\) or \(6\)

so at \(n=14\), \(2^{14}\) will have unit's digit \(4\) i.e equal to unit's digit of \(n\)

and at \(n=16\), \(2^{16}\) will have unit's digit \(6\)

Hence \(14+16=30\)

Option D

Just to clarify how I got 14 & 16, note that unit's digit is 4 when powers are 2, 6, 10 which is an AP series with a common difference of 4, power of 14 will yield unit's digit 4

Similarly, unit's digit is 6 when powers are 4,8,12 which is an AP series with a common difference of 4, power of 16 will yield unit's digit 6
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How n should be an even integer?. Is 2^n=n here?

Units digit of \(2^n\) and \(n\) is equal, the only units digit that \(2^n\) can have is 2, 4, 6, or 8 (except 1, when \(n\) is 0 - units digit don't match so this case is invalid). Now since \(2^n\) will end with 2, 4, 6 or 8, \(n\) should also end with 2, 4, 6 or 8.
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