The number of all 3-digit numbers abc (in base 10)for which
`(axxbxxc)+(axxb)+(bxxc)+(cxxa)+a+b+c=29` is

Options

A. 6
B. 10
C. 14
D. 18

Answer

C

Explanation

`abc+ab+bc+ca+a+b+c=29`
`rArrabc+ab+bc+ca+a+b+c+1=29`
`rArr(a+1)(b+1)(c+1)=30`
Product can be `2xx3xx5rArr` digits are 1, 2, 4 so possible number = 3! = 6
Product can be `1xx5xx6rArr` digits are 0, 2, 9 so possible number `=2xx2!=4`
Product can be `1xx3xx10rArr` digits are 0, 2, 9 so possible number `=2xx2!=4`
Total numbers = 14