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There are _______ three-digit numbers with digit sums of 13.
Option: 1
88

Option: 2
89

Option: 3
90

Option: 4
91

Answers (1)

best_answer

Assuming the 3-digit number is xyz.

Given that sum of the digits of the number is 13
Then $x+y+z=13$
$s+y+z=12$
Two digits can be zero, but one of them has to be greater than 1 .

Assuming that $x \geq 1, y \geq 0$ and $z \geq 0$

Again assuming that $x-1=s$
$x=1+s$
Then $1+s+y+z=13$

And
$s \geq 0, y \geq 0$ and $z \geq 0$

Then the number of solutions will be
$={ }^{12+3-1} C_{3-1}$
$= { }^{14} C_{2}$
$=91$

But $x=14$ is not possible for $s=13$
The total solutions will be $=91-1$
$=90$

Option (c) is correct.

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Rishi

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