# Show that if d0 is even, then n is even

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.
• $\sum_{j=1}^k d_jb^j=b\cdot \sum_{j=1}^kd_jb^{j-1}$ is even because it is an integer multiple of the even integer $b.$ – DanielWainfleet Mar 10 '17 at 1:03