#
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

Log in to add a comment

by udayakumar

Log in to add a comment

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

Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.

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 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