Divisible by A?

Consider the following base 4 number:

A B C D C B D A {ABCDCBDA}

where each letter represents a different digit between 0 and 3 inclusive.

Assuming A doesn't represent a zero, is the above number divisible by A?

i.e. Is A B C D C B D A A \frac{{ABCDCBDA}}{A} an integer?

Maybe Yes No

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1 solution

Geoff Pilling
Aug 15, 2017

"A" must represent 1, 2, or 3.

If it's a 1, then clearly the number divides by 1.

If it's a 2, then the number ends in a 2, which in base 4 is sufficient to conclude that the number divides by 2.

If it's a 3, then you must add up all the digits, and if it divides by 3 then the number divides by 3. This number has two 0's, two 1's, two 2's and two 3's. 0 + 0 + 1 + 1 + 2 + 2 + 3 + 3 = 12 0+0+1+1+2+2+3+3 = 12 which divides by 3.

So, y e s \boxed{yes} the number is divisible by A A .

It would be even more interesting if the base was such that the divisibility rules were different. For example using the sum of digits to get divisibility by 4 in base 5, or some similar situation.

Marta Reece - 3 years, 9 months ago

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Agreed... That would indeed be interesting... Lemme see if I can come up with something. :-)

Geoff Pilling - 3 years, 9 months ago

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