Question
$6AB5+ D58C=9351$
✓
Answer
$C + 5 = 11$
$\therefore C = 11 - 5 = 6$
and $8 + B + 1 = 15$
$\therefore B = 15 -9 = 6$
and $A + 5 + 1 = 13$
$\therefore A = 13 - 6 = 7$
and $6 + D + 1 = 9$
$\therefore D = 9 - 7 = 2$
Hence $A = 7, B = 6, C = 6$ and $D = 2$
$6765+2586=9351$
$\therefore C = 11 - 5 = 6$
and $8 + B + 1 = 15$
$\therefore B = 15 -9 = 6$
and $A + 5 + 1 = 13$
$\therefore A = 13 - 6 = 7$
and $6 + D + 1 = 9$
$\therefore D = 9 - 7 = 2$
Hence $A = 7, B = 6, C = 6$ and $D = 2$
$6765+2586=9351$
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