The units digit of the two digit integer is 3 more than the tens digit. The ratio of the product of the digits to the integer is 1/2. How do you find this integer?

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Answer 1 · 2017-08-14T17:54:17

$36$ $36$

Suppose the tens digit is $t$ $t$ .

Then the units digit is $t + 3$ $t+3$

The product of the digits is $t (t + 3) = t^{2} + 3 t$ $t(t+3) = t^2+3t$

The integer itself is $10 t + (t + 3) = 11 t + 3$ $10t+(t+3) = 11t+3$

From what we are told:

$t^{2} + 3 t = \frac{1}{2} (11 t + 3)$ $t^2+3t = 1/2(11t + 3)$

So:

$2 t^{2} + 6 t = 11 t + 3$ $2t^2+6t = 11t + 3$

So:

$0 = 2 t^{2} - 5 t - 3 = (t - 3) (2 t + 1)$ $0 = 2t^2-5t-3 = (t-3)(2t+1)$

That is:

$t = 3$ $t = 3" "$ or $t = - \frac{1}{2}$ $" "t = -1/2$

Since $t$ $t$ is supposed to be a positive integer less than $10$ $10$ , the only valid solution has $t = 3$ $t=3$ .

Then the integer itself is:

$36$ $36$