Question

Find the sum of all the four digit numbers which are divisible by 29

Answer 1

Step-by-step explanation:

Here is a brute force approach.

Let the 4 digit number be expressed as "abcd". The problem as given, then can be solved in Z29Z29 as a system of linear equations:

1000a+100b+10c+d≡14a+13b+10c+d≡01000a+100b+10c+d≡14a+13b+10c+d≡0

a+b+c+d≡0a+b+c+d≡0

If c=m,d=nc=m,d=n then, a=3m+12na=3m+12n and b=25m+16nb=25m+16n.

Now, we can enumerate the possible values of (m,n)(m,n) for which 1≤a≤91≤a≤9. The pairs are as follows:

(0,3), (0,5), (0,8), (1,0), (1,5), (2,0), (2,2), (2,5), (2,7), (3,0), (3,2), (3,7), (3,9), (4,2), (4,4), (4,7), (4,9), (5,4), (5,9), (6,1), (6,4), (6,6), (7,1), (7,6), (7,8), (8,1), (8,3), (8,6), (8,8), (9,3), (9,8)

Similarly, we can enumerate the possible values of (m,n)(m,n) for which 0≤b≤90≤b≤9. The pairs are as follows:

(0,0), (0,2), (0,4), (0,6), (1,4), (1,6), (1,8), (2,1), (2,6), (2,8), (3,1), (3,3), (3,8), (4,1), (4,3), (4,5), (4,7), (5,0), (5,5), (5,7), (5,9), (6,0), (6,2), (6,7), (6,9), (7,0), (7,2), (7,4), (7,9), (8,2), (8,4), (8,6), (8,8), (9,1), (9,6), (9,8)

If we notice, then the following pairs are common above:

(4,7), (5,9), (8,6), (8,8), (9,8)

So, there are 5 solutions: 9947, 7859, 9686, 4988, 7598.