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**Please solve in a diagrammed manner:**

$FindthevalueofA,BandCin:\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}8A\phantom{\rule{0ex}{0ex}}+8B\phantom{\rule{0ex}{0ex}}\_\_\_\_\_\_\_\_\_\phantom{\rule{0ex}{0ex}}CB3\phantom{\rule{0ex}{0ex}}\_\_\_\_\_\_\_\_\_$

$FindthevalueofA,BandCin:\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}8A\phantom{\rule{0ex}{0ex}}+8B\phantom{\rule{0ex}{0ex}}\_\_\_\_\_\_\_\_\_\phantom{\rule{0ex}{0ex}}CB3\phantom{\rule{0ex}{0ex}}\_\_\_\_\_\_\_\_\_$

We have:

8 A

+ 8 B

_____

C B 3

Since A, B and C are digits, they can only take values 0, 1, 2, 3, 4, 5, 6, 7, 8 or 9.

So, the minimum value of the sum above is 80 + 80 = 160 and the maximum value is 89 + 89 = 178. Thus, C = 1.

8 A

+ 8 B

_____

1 B 3

Now, B can either be equal to 6 or 7.

Let us assume that B = 6. Then the final sum is 163. Here, the only two numbers which add up to 163 are 82 and 81. So, 82 + 81 = 163, which gives B = 2 or 1, which is a contradiction since we had assumed B to be 6. Thus, the final sum must be 173. This gives B = 7.

So,

8 A

+ 8 7

_____

1 7 3

Now, 87 + 86 = 173, which gives A = 6.

Thus,

8 6

+ 8 7

_____

1 7 3

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