I took many hours to solve this problem given by my professor. I solve it manually like listing all possible answers but same as my earlier trials i didn't get the answer. Could someone help me with this. So much appreciated.

Okay so this is the problem...

The digits of a three-digit number are in geometric progression. If 596 is subtracted from this number ,the resulting three-digit number has its digits in arithmetic progression with a common difference equal to the reciprocal of the ratio of the geometric progression. Find the number.

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Easy Math Editor

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
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`\frac{2}{3}`	$\frac{2}{3}$
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`\sum_{i=1}^3`	$\sum_{i=1}^3$
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Comments

It is clear that the only possible cases are $111, 124, 421, 248, 842, 139$ and $931$ . Further as the number is greater than $596$ we are left with only two cases: $842$ and $931$ . Now just subtracting $596$ from each of these numbers it is clear that $842$ satisfies the given condition.

Karthik Kannan - 6 years, 11 months ago

You can also use the fact that the arithmetic progression is the reciprocal of the geometric progression ratio; that must mean that the geometric progression is decreasing, eliminating all but $3$ possibilities.

Daniel Liu - 6 years, 11 months ago

Thank you very much for your response. I also found the number 842 but i felt uncertain of answering this number because of the rule of geometric progression. Therefore i tried other numbers..

Crischell Baylon - 6 years, 11 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$