Coincidental residue powers?

Number Theory Level 3

If the last 2 digits of $\underbrace{6\times6\times 6\times \cdots \times 6}_{\text{Number of 6's } =M}$ is 76,

must the last 2 digits of $\underbrace{4\times4\times 4\times \cdots \times 4}_{\text{Number of 4's } =M}$ be 76 as well?

No Yes

1 solution

Jon Haussmann
Mar 24, 2017

$6^5 = 7776$ ends in 76, but $4^5 = 1024$ does not end in 76, so the answer is "No".

That's not true. If we go further then $6^{10} \space and 4^{10}$ hold true as per mentioned condition.

Naren Bhandari - 4 years, 2 months ago

The problem is not "can the last two digits of $4^M$ be 76 as well," the problem is " must the last two digits of $4^M$ be 76 as well". All you need is one counter-example to show that the answer is no.

Jon Haussmann - 4 years, 2 months ago

That really sucks me now When i see it is must. Thank!

Naren Bhandari - 4 years, 2 months ago

Coincidental residue powers?

1 solution

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