solve the 10th sum
Dear Student,
Let a = the hundreds digit, b = the tens digit, c = the units digit.
Let a = the hundreds digit, b = the tens digit, c = the units digit.
In a 3 digit number, unit digit is more than the hundred digit,
c > a
Unit's digit, Ten's digit, Hundred's digit are in a ratio 1:2:3
b = 2a
c = 3a
Original number = 100c+10b+a
Number obtained by reversing digits = 100a+10b+c
if the difference of original number and the number obtained by reversing the digit is 594,
Reverse number - original number 594
(100c + 10b + a) - (100a + 10b + c) = 594
100c + 10b + a - 100a - 10b - c = 594
99c - 99a = 594
simplify, divide by 99
c - a = 6
we know c = 3a
3a - a = 6
2a = 6
a = 3
then
b = 2(3) = 6
c = 3(3) = 9
Hence, the original number is 369
Regards c > a
Unit's digit, Ten's digit, Hundred's digit are in a ratio 1:2:3
b = 2a
c = 3a
Original number = 100c+10b+a
Number obtained by reversing digits = 100a+10b+c
if the difference of original number and the number obtained by reversing the digit is 594,
Reverse number - original number 594
(100c + 10b + a) - (100a + 10b + c) = 594
100c + 10b + a - 100a - 10b - c = 594
99c - 99a = 594
simplify, divide by 99
c - a = 6
we know c = 3a
3a - a = 6
2a = 6
a = 3
then
b = 2(3) = 6
c = 3(3) = 9
Hence, the original number is 369