I recently came across this question:

Suppose $b ∈ N$ is even. Let $$ n = (d_{k} d_{k-1} . . . d_{1} d_{0} )_{b} = d_{k} b^{k} + d_{k-1} b^{k-1} + · · · + d_{1} b + d_{0} $$

Show that if $d_{0}$ is even, then $n$ is even.

I remember the basic rules: $n$ is even if $n=2k$ for some integer $k$, and odd if $n=2l + 1$ for some integer $k$. The question seems easy enough but I can't show what the question asks for.

I tried seeing if the converse is true and maybe working from there but I can't figure it out from that angle either.

I also considered assigning $d_{0}$, $0$ and assigning $k$ and $b$ a random but easy to compute value but that doesn't seem right at all.

I'd really appreciate some help with this. Thank you!


Since $b$ is even, all its powers are also even, i.e. $$ b,b^2,\ldots, b^k $$ are all even. Therefore $$ m=d_kb^k+d_{k-1}b^{k-1}+\ldots+d_2b^2+d_1b $$ is even. Hence, $$ n=d_kb^k+d_{k-1}b^{k-1}+\ldots+d_2b^2+d_1b+d_0=m+d_0 $$ is even if and only if $d_0$ is also even.

  • $\begingroup$ $\sum_{j=1}^k d_jb^j=b\cdot \sum_{j=1}^kd_jb^{j-1}$ is even because it is an integer multiple of the even integer $ b. $ $\endgroup$ – DanielWainfleet Mar 10 '17 at 1:03
  • $\begingroup$ @user254665 I don't get it? Isn't a finite sum of even integer an even integer? $\endgroup$ – Mercy King Mar 10 '17 at 4:42
  • $\begingroup$ Thank you. I really appreciate it! $\endgroup$ – Drew U Mar 10 '17 at 5:14
  • $\begingroup$ I just wanted to point out a different way. Nothing wrong with your answer. I gave it a +1. $\endgroup$ – DanielWainfleet Mar 10 '17 at 5:23

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