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.

## Comments

Sort by:

TopNewestIt 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 · 3 years, 2 months ago

Log in to reply

– Daniel Liu · 3 years, 2 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.Log in to reply

– Crischell Baylon · 3 years, 2 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..Log in to reply