Question: The square of the sum of the digits in a two-digit number is 64. If 3 is added to four times the number, the digits are reversed. Find the number.

Solution:

Let the digit at tens place of a two-digit number be x and that at ones place be y. So, the number is 10x +y.

According to the question,

Condition I,
The square of the sum of the digits is 64.
$or, (x+y)² = 64$
$or, (x + y)² = 8²
$or, x + y = 8$ - (i)

Condition II,
If 3 is added to four times the number, the digits are reversed.
$or, 3 + 4(10x +y) = 10y +x$
$or, 3 +40x +4y = 10y +x$
$or, 3 +40x -x = 10y -4y $
$or, 3 + 39x = 6y$
$or, 3(1+ 13x) = 3(2y)$
$or, 1 + 13x = 2y$
$or, y = \frac{1 +13x}{2}$ - (ii)
 
Put value of y from equation (ii) in equation (i), we get,

$or, x + \dfrac{1 +13x}{2} = 8$

$or, \dfrac{2x + 1 + 13x}{2} = 8$

$or, 15x +1 = 16$

$or, 15x = 16-1$

$or, 15x = 15$

$\therefore x = 1$

Put value of x in equation (ii), we get,

$or, y = \frac{1 + 13×1}{2}$
$or, y = \frac{14}{2}$
$\therefore y = 7$

So, (x,y) = (1,7)
10x = 10×1 = 10
10x + y = 10 +7 = 17

Therefore, the required two digit number is 17.

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