Question

The sum of digits of a two digit number is 13. If sum of their squares is 89, find the
number.​

Answer 1

$Let \: x \:and \: ( 13 - x ) \:are \: digits \: in \:a \\two \: digit \: number$

$The \: number = 10x + 13 - x \\= 9x + 13 \: ---(1)$

/* According to the problem given */

$x^{2} + ( 13 - x )^{2} = 89$

$\implies x^{2} + 13^{2} - 2 \times 13 \times x + x^{2} = 89$

$\implies x^{2} + 169 - 26x + x^{2} - 89 = 0$

$\implies 2x^{2} - 26x + 80 = 0$

/* Divide each term by 2 , we get */

$\implies x^{2} - 13x + 40 = 0$

/* Splitting the middle term,we get */

$\implies x^{2} - 5x - 8x + 40 = 0$

$\implies x( x - 5 ) - 8( x - 5 ) = 0$

$\implies ( x - 5 )( x - 8 ) = 0$

$\implies x - 5 = 0\: Or \: x - 8 = 0$

$\implies x = 5 \: Or \: x = 8$

Case 1 :

$If \: x = 5 , then \\ the \: number \\= 9x +13 \\= 9 \times 5 + 13 \\= 45 + 13 \\= 58$

Case 2:

$If \: x = 8 , then \\ the \: number \\= 9x +13 \\= 9 \times 8 + 13 \\= 72 + 13 \\= 85$

Therefore.,

$\red { Required \: number \: is } \green { = 58 \: Or \: 85 }$

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