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Number Theory

A whole number \(n\) is divisible by \(2\) if and only if the last digit of the decimal representation of \(n\) is divisible by \(2.\)

Proof \(\downarrow\)

Let \(n\) be a whole number and let the decimal representation of \(n\) be \[n = d_k d_{k-1} \dots d_2 d_1\] Expand the number at the one's place. \[d_k d_{k-1} \dots d_2 \cdot 10 + d_1\] \[= 2 \cdot k + d_1\] where \(k\) represents the integer \(d_k d_k-1 \dots d_2 \cdot 5.\)

If \(d_1\) is a multiple of \(2\) then \(d_1 = 2 \cdot j\) for some integer \(j,\) in which case \(n = 2 \cdot (k + j)\) and is divisible by \(2.\) On the other hand, if \(d_1\) is odd then \(d_1 = 2 \cdot j + 1\) for some integer \(j,\) in which case \(n = 2 \cdot (k + j) + 1\) is odd and not divisible by \(2.\)