Calling All Powers That Give Two Digits

Number Theory Level 2

Let $A,B,C$ and $D$ be distinct single digit positive integers such that: $A^B =\overline{CD} \text{ and } A+B=C+D$

Compute $\overline{AB} + \overline{BC} + \overline{CD} + \overline{DA}$ .

Clarification : $\overline{AB}$ denotes a 2-digit integer with $A$ and $B$ as its digits.

The answer is 220.

1 solution

Zeeshan Ali
Feb 10, 2016

$\huge{\color{#20A900}{8} \neq \color{#3D99F6}{2} \neq \color{#69047E}{6} \neq \color{#624F41}{4}}$ $\huge{\color{#20A900}{8} ^ \color{#3D99F6}{2} = \color{#69047E}{6} \color{#624F41}{4}}$ $and$ $\huge{\color{#20A900}{8} + \color{#3D99F6}{2} = \color{#69047E}{6} + \color{#624F41}{4}}$ $therefore$ $\huge{\color{#20A900}{8} \color{#3D99F6}{2} + \color{#3D99F6}{2} \color{#69047E}{6} + \color{#69047E}{6} \color{#624F41}{4} + \color{#624F41}{4}} \color{#20A900}{8} \, = \, \boxed{\color{#D61F06}{220}}$

I have the same solution. But how can we guarantee that there's a unique solution to this problem?

Nihar Mahajan - 5 years, 4 months ago

You can try all the possible combinations, they are few! :)

Zeeshan Ali - 5 years, 4 months ago

How about 5^2 ?

Irvine Dwicahya - 5 years, 3 months ago

@Irvine Dwicahya – In that case we do not have all distinct digits :)

Zeeshan Ali - 5 years, 3 months ago

Calling All Powers That Give Two Digits

The answer is 220.

1 solution

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