A number consists of two digits. The digit in the ten's place is greater than unit place by 4. sum of square of digits of the number is 15 less than the number. Find the number.

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Solution

Let unit's digit be x.
Then ten's digit = $x + 4$
Number = $10 (x + 4) + x = 11 x + 40$
Therefore,
$x^{2} + (x + 4)^{2} + 15 = 11 x + 40$
$2 x^{2} - 3 x - 9 = 0$
$(x - 3) (2 x + 3) = 0$
$x = 3$
Hence, the number is $33 + 40 = 73$ .

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A number consists of two digits. The digit in the ten's place is greater than unit place by 4. sum of square of digits of the number is 15 less than the number. Find the number.

Let unit's digit be x.Then ten's digit = x+4Number = 10(x+4)+x=11x+40Therefore,x2+(x+4)2+15=11x+402x2−3x−9=0(x−3)(2x+3)=0x=3Hence, the number is 33+40=73.

Let unit's digit be x.
Then ten's digit = $x + 4$
Number = $10 (x + 4) + x = 11 x + 40$
Therefore,
$x^{2} + (x + 4)^{2} + 15 = 11 x + 40$
$2 x^{2} - 3 x - 9 = 0$
$(x - 3) (2 x + 3) = 0$
$x = 3$
Hence, the number is $33 + 40 = 73$ .