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Question

If abc-bca = 792, where abc is a 3-digit number, if a+b+c = 18 and a = 9c, what is the number?

Solution
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We have, abccba=792
abc is a 3digit number.
Hence, abc=100a+10b+c
Similarly, cba=100c+10b+a
So, abcbca=100a+10b+c100c10ba
=99a99c
=99(ac)

So, 99(ac)=792
ac=8
Also, a=9c[given]
9cc=8
8c=8
Hence, c=1

So, a=9(1)
Hence, a=9

Also, we are given, a+b+c=18
So, 9+b+1=18
b=8

Hence, the number is 981.

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