Question

the product of 2 degit numbers is 1998 if the product of their units degits is 28 and that of tens degits is 15,find the numbers????????

Answer 1

The required two numbers are 37 and 54

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Answer 2

$let \: the \: numbers \: be \: 10x + y \: and \: 10a + b \\ given \: (10x + y)(10a + b) = 1998 \\ and \: by = 28 \\ ax = 15 \\ 100ax + yb + 10xb + 10ay = 1998 \\ 1528 + 10(xb + ay) = 1998 \\ xb + ay = 47 \\ now \: b = \frac{28}{y} \\ a = \frac{15}{x} \\ 28 \frac{x}{y} + 15 \frac{y}{x} = 47 \\ let \: \frac{x}{y} = t \\ 28 {t}^{2} - 47t + 15 = 0 \\on \: solving \\ t = \frac{5}{4} \\ t = \frac{3}{7} \\ now \\ \frac{ax}{by } = \frac{15}{28} \\ \frac{at}{b} = \frac{15}{28 } \\ first \: case \: \\ t = \frac{5}{4} \\ = > \frac{a}{b} = \frac{3}{7} \\ now \: let \: the \: proportion \: factor \: be \: 1 \\ then \\ \: x = 5 \\ \: y = 4 \: \\ a = 3 \: \\ b = 7 \\ now \: the \: numbers \: are \: 5(10) + 4 \: and \: 3(10) + 7 \\ i.e. \: 54 \: and \: 37 \\ second \: case \: \\ t = \frac{3}{7} \\ = > \frac{a}{b} = \frac{5}{4} \\ now \: let \: the \: proportion \: factor \: be \: 1 \\ then \\ \: x = 3 \: \\ y = 7 \: \\ a = 5 \: \\ b = 4\\ now \: the \: numbers \: are \: \: 3(10) + 7 \: and \: 5(10) + 4 \\ i.e. \: 37 \: and \: 54 \\ verification > \\ 37 \times 54 \\ = 1998 \\ hence \: verified \: \\ the \: numbers \: are \: 37 \: and \: 54$
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