Question #322471

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.


1
Expert's answer
2022-04-05T02:35:52-0400

Let the unknown number be 100x+10y+z100x+10y+z


The sum of the digits

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


100z+10y+x=594+100x+10y+z100z+10y+x=594+100x+10y+z


=>99z99x=594=> 99z-99x=594


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


100x+10z+y=36+100x+10y+z100x+10z+y=36+100x+10y+z


=>9z9y=36=> 9z-9y=36


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



From (2), we have that x=z6x=z-6


=>x+y+z=z6+y+z=11=> x+y+z=z-6+y+z=11


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



From (3), we have that y=z4y=z-4


=>2z+z4=17=> 2z+z-4=17

=>3z=21=>3z=21


Hence, z=7z=7



y=z4=74=3y=z-4=7-4=3


=>y=3=> y=3



x=z6=76=1x=z-6=7-6=1


=>x=1=> x=1



Thus, the number is 100(1)+10(3)+7=100+30+7=137100(1)+10(3)+7=100+30+7=137



Original number is 137137





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