Solution to "Sum of Digits"

View Problem: Sum of Digits

The value of the number abc is equal to 100*a + 10*b + c. Similarly, the value of def is equal to 100*d + 10*e + f. Adding these two gives us 100*(a+d) + 10*(b+e) + (c+f), which is the value of ghij. Let us try to represent g, h, i and j in terms of a, b, c, d, e and f:

Case 1: (a+d) < 10, (b+e) < 10, (c+f) < 10

For this case,

• g = 0
• h = a+d
• i = b+e
• j = c+f

So, g + h + i + j = a + b + c + d + e + f. Hence (a + b + c + d + e + f) % 9 = (g + h + i + j)%9

Case 2: (a+d) >= 10, (b+e) < 10, (c+f) < 10

For this case,

• g = 1
• h = (a+d)%10
• i = b+e
• j = c+f

If we prove that (a + b + c + d + e + f - (g + h + i + j)) % 9 = 0, we are done.

  (a + b + c + d + e + f - (g + h + i + j)) % 9 
= (a + b + c + d + e + f - (1 + (a+d)%10 + b+e + c+f)) % 9
= (a + d - 1 - (a+d)%10) % 9
= (10-1) % 9 //(a+d)-(a+d)%10 = 10 as a, d lie in [0..9] so (a+d) lies in [0..18]
= 9 % 9
= 0

Hence proved for this case.

Similarly, the proof can be continued for all cases, and generalized as well.

Sum of Digits

Say we have 2 numbers abc and def, where, a, b, c, d, e and f are the digits of the 2 numbers lying in [0..9]. Adding these 2 numbers gives us the number ghij. Again, g, h, i and j are the digits of the resulting number.

  abc
+ def
----------
 ghij

Can you prove that

(a + b + c + d + e + f) % 9 = (g + h + i + j) % 9 

is always true?

To understand better, let us see an example -

  379
+ 822
----------
 1201

(3 + 7 + 9 + 8 + 2 + 2) % 9 = 31 % 9 = 4
(1 + 2 + 0 + 1) % 9 = 4 % 9 = 4

Let us now generalize the above statement. Say we have N numbers A₁, A₂, ... A_N. Let S(x) denote the sum of digits of the the number x.

Can you prove that

[S(A₁) + S(A₂) + .. + S(A_N)] % 9 = S(A₁ + A₂ + .. + A_N) % 9

is always true?

To put it in a single sentence, can you prove that the sum of digits of a group of numbers modulo 9 is the same as the sum of digits of the sum of those numbers modulo 9 ?

View Solution: Solution to "Sum of Digits"