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I have to find a $6$ digit integer, greater than $800000$ and divisible by $4$, where the same $3$ digits appear twice each and add to $24$. I do not even know where to start from!

1 Answers 1

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  • "The three numbers add to $24$", "Same 3 numbers twice (each)"

If we assume this means that the three numbers must be different, we notice first that $9+8+7=24$ is the largest possible sum of the three different one-digit numbers and if we were to replace any of these numbers with something smaller, the total would be strictly less than $24$.

  • "The number is divisible by $4$"

This implies the final two digits must be a two-digit number divisible by four. This implies that the final digit must be an $8$ in order to ensure that it is even. Since $98$ and $78$ are both not divisible by four the only choice is $88$

  • "The number is greater than $800000$", "six digit whole number"

Since both $8$'s must be at the end and any six digit number starting with a seven is less than $800000$, the first digit must be a $9$

The number is then $9\star\star\star88$ where among the $\star$'s one more nine and two sevens are placed.

$997788, 977988, 979788$