A change of base

Number Theory Level 2

This is a 6-digit number in base 7:

$\large aabccb_7$

If $a$ , $b$ , and $c$ represent 3 distinct digits (from 0 to 6), is $aabccb_7$ odd or even?

Not enough information Odd Even

2 solutions

David Vreken
Nov 10, 2018

$aabccb_7$ is equal to $7^5a + 7^4a + 7^3b + 7^2c + 7c + b$ or $(7^5 + 7^4)a + (7^3 + 1)b + (7^2 + 7)c$ in base $10$ .

Since $7^m$ is odd (for all integers $m \geq 0$ ), $7^m + 7^n$ is even (for all integers $m \geq 0$ and $n \geq 0$ ). This makes each coefficient of $a$ , $b$ , and $c$ even, which makes the whole expression always even .

Geoff Pilling
Nov 9, 2018

In base 7, even-ness is determined by adding up the digits. Since each digit is in there twice, the sum is clearly even, so the number is even as well.

Does this work for all odd bases? I think it does, but I'm not quite sure.

Henry U - 2 years, 7 months ago

Yes, it does.

Every digit represents a multiple of a power of $7$ , this power is always odd and therefore the digit represents a numbet of the same parity as this digit itself, so adding up all digits gives the same parity as adding up the numbers they represent.

$\begin{aligned} n_{2k+1} &= \displaystyle \sum_{i=0}^{\infty} d_i \cdot (2k+1)^i \\ &\equiv \sum_{i=0}^{\infty} d_i \pmod 2 \end{aligned}$

Henry U - 2 years, 7 months ago

A change of base

2 solutions

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