Question

The number of all 3-digit numbers abc (in base 10) for which $(a\times b\times c)+b(a+c)+(c\times a)+a+b+c=29$ is

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

Answer 1

Answer: (C)

$14$

We have,

$abc+ab+a+ca+bc+b+c+1=30ab(c+1)+a(c+1)+b(c+1)+c+1=30$

$(c+1)(ab+a+b+1)=30(c+1)(b+1)(a+1)=30$

Now, $30$ can be expressed as product of three nos $(x,y,z)$ in following ways:
[such that $x-1,y-1,z-1$ belong to set $\{0,1,2,3,4,5,6,7,8,9\}$]

case 1: $x,y,z\in \{2,3,5\}$

case 2: $x,y,z\in \{1,5,6\}$

case 3: $x,y,z\in \{1,3,10\}$

For Case 1: The digits can be assigned values in $3!$ ways.
For Case 2: Since, a in abc cannot be 0 [as the no will become a 2-digit no] there we have only two places to assign 1 ie. ones and tens digit. and then $2\times 1$ for the remaining two digits. Thus, $2\times 2\times 1=4$ ways for case #2.
Similarly, for Case 3: we have $2\times 2\times 1=4$ (Same case as Case #2).
So, the total number of all 3 digit numbers is $14$