# Cryptarithmetic Problem- 5

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**Multiplication**

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Why can E not be 5?

ReplyDeleteIf A+E+carry =Q,without generating any carry over,then how can we conclude that E != 5.

ReplyDeleteplease reply.looking forward for ur ans

2+5+carry (2) =9 no carry over

ReplyDelete3+5+carry (1) =9 no carry over

1+5+carry (2) =8 no carry over

1+5+carry (1) =7 no carry over

As A can only be 2, 4, 6, 8 (since A+2B = 20 0r 10 from third column from left, K=0) Why cant A=2 and E =5

ReplyDeleteThe value of Q will be 7.

ReplyDeletethe value of Q will be 7

ReplyDeleteThe approach to find the value of p i.e 5 is not good since there are less than ten variables so p can have other value as well.

ReplyDeletebetter approach is to first check the following multiplication

ReplyDeletePAS* B=ASAA

try to find answer from here. If you could not i will help u.

here Q=7 as R&Q both cant be 6.

ReplyDelete