Question

What least number must be subtracted from each of the numbers 7, 17 and 47 so that the remainders are in continued proportion?

Answer 1

let x be the number that need to be subtracted from these numbers to make it proportional.

The new numbers will be $7-x,17-x,47-x$

Since these numbers should be proportional,

$\frac{17-x}{7-x}=\frac{47-x}{17-x}$

By simplification we get,

$(17-x)(17-x)=(47-x)(7-x)$

$x^2-34x+289=329+x^2-54x$

$20x=40\Rightarrow x=2$

The new numbers will be $5,15,45$

${}^{\prime }2{}^{\prime }$ is the least number to be subtracted so that the numbers will be in proportional.