Question

Find out (A+B+C+D) such that AB×CB=DDD,where AB and CB are two digit numbers DDD is three digit-number?

$\huge answer \: \mapsto 21.$
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Answer 1

$➪ \: AB×CB=DDD$

➪ Now DDD has a factor of D or 111

➪ 111 has factor 3 or 37

$➪ \: DDD=37×3×D \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: =AB×CB$

➪ Thus either AB and CB is 37 and other number is 3\times D.

$➪ \: D=9$

$➪ \: 37×(3×9)=999$

$➪ \: 37×27=999$

$➪ \: A=3,B=7,C=2,d=9$

$➪ \: A+B+C+D \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: =3+7+2+9 \\ \:=21$

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Answer 2

Step-by-step explanation:

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