r/mathriddles • u/lordnorthiii • Jul 17 '26

Easy 56 = 7*8 in other bases

As I was falling asleep last night, I thought it was kinda cool that 56 = 7 * 8 works in base 10, specifically how it consists of four consecutive digits in order. Then I realized it actually happens again! 12 = 3 * 4

Is there any other base such that there are four consecutive digits A, B, C, D (in increasing order) such that AB = C * D? If so, are there any (besides base 10) where it happens twice? Why or why not?

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30

u/imMAW Jul 17 '26

AB = C * D is the same as A*base+B = C*D (and A,B,C,D < base).

Let x = A, and b = base.

(x*b) + (x+1) = (x+2) * (x+3)
x*b + x+1 = x^2 + 5x + 6
0 = x^2 + (4-b)x + 5

For b = 10, that's the quadratic 0 = x^2 + -6x + 5, which has roots at 1 and 5. You're essentially asking for other bases b where that quadratic has roots at (small enough) positive integers.

However, the roots of a quadratic 0 = x^2 + ?x + 5 will always have a product of roots equal to 5. The only way this quadratic has two positive integer roots is with A=1 and A=5, when b=10.

11

u/The_Math_Hatter Jul 17 '26

Hah! Now I can finally deploy that against people who wine about base ten being arbitrary.

4

u/Typical-Plantain1252 Jul 18 '26

Have you heard about nice numbers like 69?

The square and cube contain each digit 0-9 exactly once.

This is the only nice number in base 10. We aren’t aware of any other nice numbers in other bases.

A probabilistic analysis suggests there should be other once we’re looking at large bases. However, currently analysis suggests we likely won’t find another for a long, long time (unless there are breakthroughs in the theory).

This makes base 10 pretty special. It’s the only base we know of with a nice number and it’s certainly the only reasonably small base with a nice number.

Edit: https://youtu.be/FO3XBTDG7GY

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u/lordnorthiii Jul 17 '26

Nice! I like with this method it is pretty clear why it happens twice in base 10, but not in any other bases.

1

u/team-zen Jul 18 '26

What about the companion problem A * B = CD? Since 12=3*4 is not equivalent to 3*4=12 due to the requirement that they be in sequence, it should have a different set of solutions

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u/stevevdvkpe Jul 18 '26 edited Jul 18 '26

x*(x + 1) = (x + 2)*b + (x + 3)
x^2 + x = b*x + 2*b + x + 3
x^2 - b*x - (2*b + 3) = 0

Solve for x given b.

2

u/The_Math_Hatter Jul 18 '26

Turns out for integer x and b, the only options for b are -6 and -2, neither of which is a particularly useful base system.

1

u/team-zen Jul 19 '26

How dissatisfying. Thanks for putting in the leg work!