Question

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The units and tens digits of one-two digit integer are the tens and units digits of another two- digit integer, respectively. If the product of the two integers is 4930, what is their sum?

Answer 1

(10t + u)(10u + t) = 4,930

100tu + 10t^2 + 10u^2 + tu = 4,930

101tu + 10t^2 + 10u^2 = 4,930

10t^2 + (101u)t + (10u^2 - 4,930) = 0

(101u)^2 - 4(10)(10u^2 - 4,930)

= 10,201u^2 - 400u^2 + 197,200

= 9,801u^2 + 197,200

We see that u = 5 gives 442,225.

√442,225 = 665

(10t + 5)(50 + t) = 4,930

500t + 10t^2 + 250 + 5t = 4,930

10t^2 + 505t - 4,680 = 0

2t^2 + 101t - 936 = 0

t = (-101 + √(101^2 - 4(2)(-936)))/(2×2)

= (-101 + √17,689)/4

= (-101 + 133)/4 = 32/4 = 8

So the numbers are 58 and 85.

58 + 85 = 143