# A,b,c are digits of a three-digit number such that 64a+8b+c=403,then the value of a+b+c+2013 is?

2
by udayakumar

• Brainly User
2014-09-13T11:50:52+05:30
64a+8b+c=403 = 400+3
Therefore c has to be =3.
Now divide 400 by 64 until remainder is multiple of 8 with a single digit number.
You will see that 400 = 64x6 + 8x2.
Hence b = 2 and a = 6.
Hence a+b+c+2013 = 6+2+3+2013 = 2024
2014-09-15T13:16:13+05:30

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Its a bit logical as well as mathemetical
64a+8b cannot be odd so
64a+8b+c=403 = 400+3
so c =3
so
400 = 64 x 6 + 8 x 2
comparing with 64a + 8b
we have a = 6 ,b =  2
so
a + b + c + 2013 = 6 + 2 + 3 + 2013 = 2024
64a+8b can be even and can be equal to 402. why can't it be ? or can it be?
64a+8b cannot be odd as when we multiply any number to an even number we get a even number so here a,b can be any number(even or odd) but as it is multiplied by even number 64 and 8 respectively so the resultant(sum iof two even) will also be a even.
64a+8b = 8(8a+b), so whatever a and b, 64a+8b has to be divisible by 8 which 402 is not.