Answer to Question #322471 in Linear Algebra for Kenth Smith

Question

The sum of the digits of a three-digit number is 11. If the first and last digits are interchanged, the new number is greater than the original number by 594. If the tens and the units digits are interchanged, the resulting number is greater than the original by 36. Find the original number.

Answer 1

Let the unknown number be $100x+10y+z$

The sum of the digits

$x+y+z=11$ ---------->(1)

$100z+10y+x=594+100x+10y+z$

$=> 99z-99x=594$

$=> z-x=6$ -------------->(2)

$100x+10z+y=36+100x+10y+z$

$=> 9z-9y=36$

$=> z-y=4$ --------------->(3)

From (2), we have that $x=z-6$

$=> x+y+z=z-6+y+z=11$

$=> 2z+y=17$ --------->(4)

From (3), we have that $y=z-4$

$=> 2z+z-4=17$

$=>3z=21$

Hence, $z=7$

$y=z-4=7-4=3$

$=> y=3$

$x=z-6=7-6=1$

$=> x=1$

Thus, the number is $100(1)+10(3)+7=100+30+7=137$

Original number is $137$