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# 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.

Note by Crischell Baylon
2 years, 3 months ago

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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. · 2 years, 3 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. · 2 years, 3 months ago