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

Hint: In this problem based on the given expression using the variables a, b, c as of the form \[(a \times b \times c) + b(a + c) + (c \times a) + a + b + c = 29\] and we need to solve the expression given using the variables assigned and using the factorial formulas we could able to find the answer.

Complete step-by-step answer:
We are given the expression is,
\[(a \times b \times c) + b(a + c) + (c \times a) + a + b + c = 29\]
It can be written as,
\[(abc) + (ab + bc) + (ca) + a + b + c = 29\]
\[abc + ab + bc + ca + a + b + c = 29\]
Adding \[1\] on both the sides of the expression then it comes as follows,
\[abc + ab + bc + ca + a + b + c + 1 = 29 + 1\]
\[abc + ab + bc + ca + a + b + c + 1 = 30\]
Grouping and making the common terms to be out of it then,
The orders arranged to make it into the group of variables
\[abc + ab + a + ca + bc + b + c + 1 = 30\]
Among the first two terms \[ab\] is common and in next two terms \[a\] is common and in next two terms \[b\] is common thus among all \[(c + 1)\] is made to common,
\[ab(c + 1) + a(c + 1) + b(c + 1) + c + 1 = 30\]
\[(c + 1)(ab + a + b + 1) = 30\]
And with the upcoming \[(ab + a + b + 1)\] in this first two terms \[a\] is common then,
\[(c + 1)(a(b + 1) + (b + 1)) = 30\]
\[(c + 1)(b + 1)(a + 1) = 30\]
Thus, the expression is made to be simplified.
Now, RHS \[30\] can be expressed as product of three numbers \[(x,y,z)\] in following ways:
Factors of \[30\] are - \[\{ 1,2,3,5,6,10\} \]
[such that \[x - 1,y - 1,z - 1\]belong to set \[\{ 0,1,2,3,4,5,6,7,8,9,\} \]]
Case 1:
Let the set for \[(x,y,z)\] belongs to \[\{ 2,3,5\} \]
The digits can be assigned values in \[3!\] ways.
Case 2:
Let the set for \[(x,y,z)\] belongs to \[\{ 1,5,6\} \]
Since \[a\], in \[abc\] cannot be \[0\] [as the number will become a two-digit number] there we have only two places to assign \[1\].
That is, ones and tens digits.
And then \[2 \times 1\] for the remaining two digits.
Thus, \[2 \times 2 \times 1\] ways for this case = \[4\]
Case 3:
Let the set for \[(x,y,z)\] belongs to \[\{ 1,3,10\} \]
Which is similar to case 2 and hence the ways are \[2 \times 2 \times 1\]= \[4\].
Hence the total number of \[3\] digit numbers such that for the expression is
\[(a \times b \times c) + b(a + c) + (c \times a) + a + b + c = 29\]
\[ \Rightarrow 3! + 4 + 4\]
\[
   \Rightarrow (3 \times 2 \times 1) + 4 + 4 \\
   \Rightarrow 6 + 4 + 4 \\
   \Rightarrow 14 \;
 \]
Hence the total number of three-digit numbers is \[14\] numbers.
Based on the option given, Option C is the correct one.
So, the correct answer is “Option C”.

Note: Interesting facts about factors of the number as factors play the important role in this sum,
I.Every digit is a factor in and of itself.
II.Every number's component is either less than or equal to that number.
III.Every number has a factor of one.
IV.Every divisor of a given number is also a factor of that number.
V.A given number has a finite number of factors.