Sum Of Three-digit Number Is 17. Sum Of Squared Of Digits Of The Give

1.	Sum Of Three-digit Number Is 17. Sum Of Squared Of Digits Of The Given Num Is 109. If We Subtract 495 From That Num We Will Get A Number Written In Square Order. Find The Num?
Answer» Let the nos be: abc As sum of the digit is 17. Therefore, a+b+c=17----(1) Also sum of SQUARE of digits is 109 i.e a^2+b^2+c^2=109----(2) Also, (100a+10b+c) – 495 = (100C+10b+a) or, (100a – a) + (10b – 10b) + (c – 100c) = 495 or, 99 (a-c) =495 or (a - c) = 495 The possible COMBINATIONS are (6,1) (7,2) (8,3), (9,4) For 1st combination (6,1); b = (17 – 6 - 1) = 10 which is not possible For 2nd combination (7,2); b = (17 – 7 - 2) = 8 but a^2+b^2+c^2 =! 109 so not possible For 3rd combination (8,3) ; b = (17 – 8 - 3) = 6 also a^2+b^2+c^2 = 109 so it is possible so,863 is the answer. Let the nos be: abc As sum of the digit is 17. Therefore, a+b+c=17----(1) Also sum of square of digits is 109 i.e a^2+b^2+c^2=109----(2) Also, (100a+10b+c) – 495 = (100c+10b+a) or, (100a – a) + (10b – 10b) + (c – 100c) = 495 or, 99 (a-c) =495 or (a - c) = 495 The possible combinations are (6,1) (7,2) (8,3), (9,4) For 1st combination (6,1); b = (17 – 6 - 1) = 10 which is not possible For 2nd combination (7,2); b = (17 – 7 - 2) = 8 but a^2+b^2+c^2 =! 109 so not possible For 3rd combination (8,3) ; b = (17 – 8 - 3) = 6 also a^2+b^2+c^2 = 109 so it is possible so,863 is the answer.

Sum Of Three-digit Number Is 17. Sum Of Squared Of Digits Of The Given Num Is 109. If We Subtract 495 From That Num We Will Get A Number Written In Square Order. Find The Num?

Answer»

Let the nos be: abc

As sum of the digit is 17. Therefore, a+b+c=17----(1)

Also sum of SQUARE of digits is 109 i.e a^2+b^2+c^2=109----(2)

Also, (100a+10b+c) – 495 = (100C+10b+a)

or, (100a – a) + (10b – 10b) + (c – 100c) = 495

or, 99 (a-c) =495 or (a - c) = 495

The possible COMBINATIONS are (6,1) (7,2) (8,3), (9,4)

For 1st combination (6,1); b = (17 – 6 - 1) = 10 which is not possible

For 2nd combination (7,2); b = (17 – 7 - 2) = 8 but a^2+b^2+c^2 =! 109 so not possible

For 3rd combination (8,3) ; b = (17 – 8 - 3) = 6 also a^2+b^2+c^2 = 109 so it is possible

so,863 is the answer.