A number theory problem by suresh jh

If 64a+8b+c =403 , c=3 because rest is multiple of 8.

Divide (403-3=400) by 8, then it 8a+b=50 will appear, b=2,because rest is multiple of 8, and now we can find,"a" which is 6.

Now we know a=6,b=2,c=3, then a+b+c = 11.

$\text{64(6)+8(2)+(3)}$ , $\text{6+2+3 = (11)}$

$\text{64 is an even number thus 64(a) must be even.}$

$\text{8 is an even number thus 8(b) must be even.}$

$\text{(even × even = even) because (403) is an odd thus (c = must be odd).}$

The most possible value for number that we can make with equation is $\text{64(6)+8(2) = 400}$ .

$\text{403 - 400 = 3}$ thus $\text{C is (3)}$ .

$\therefore$ the answer is $\boxed{\color{#20A900}{11}}$