Question

How many positive integers less than 1000 are 6 times the sum of their digits?

A) 0

B) 1

C) 2

D) 4

Answer 1

Hint:  We’ll suppose a number $100a + 10b + c$, where $a,b,c \in \{ 0,1,2,3.....9\} $, then we’ll create the equation applying the condition that the number is 6 times the sum of its digit.

Then we will solve the equation for a,b and c to get the required numbers.

Complete solution:

Given, the number is 6 times the sum of their digits.

Let a number be $100a + 10b + c$, where $a,b,c \in \{ 0,1,2,3.....9\} $

If \[a = b = 0\]and \[c \ne 0\]then it’ll be a single-digit positive number

If \[a = 0\] and \[b \ne 0\] then it’ll be a two-digit number

If \[a \ne 0,\] then it’ll be a three-digit number

Now, following the condition given we’ll have an equation

i.e. $100a + 10b + c = 6(a + b + c)$

now, we’ll put together the like terms to simplify the

$ \Rightarrow 100a - 6a + 10b - 6b = 6c - c$

$ \Rightarrow 94a + 4b = 5c...........(i)$

To get the answer of equation(i), \[a = 0\] as if ‘a’ will have any value aside from ‘0’, no value of ‘c’ will satisfy equation(i). Since ‘c’ can have a maximum value as 9 which makes RHS 45 and it is less than 94 in any case of RHS.

Therefore, \[a = 0\]

$4b = 5c$, that has just one solution,

$\therefore b = 5$

And $c = 4$

Therefore, there’s just one positive value less than 1000 and six times the sum of their digits i.e.,

${\text{ = 100(0) + 10(5) + 4}}$

${\text{ = 50 + 4}}$

${\text{ = 54}}$

Hence, there is only one possible solution.

(B) is the correct option.

Note: Another technique for this question will be

For a single-digit positive number, it’s out of the question to be six times of itself.

For two-digit number,

Let the number be ${\text{10a + b}}$ where $a \in \{ 1,2,3.....9\} \;$ and $\;b \in (0,1,2,3.....9\} $

According to the condition given,

i.e. $10a + b = 6(a + b)$

$ \Rightarrow 4a = 5b$

which has just one resolution,

a = 5, b = 4

For 3-digit number,

Let the number be $100a + 10b + c$ where $a \in \{ 1,2,3.....9\} \;$ and $b,c \in \{ 0,1,2,3.....9\} $

According to the condition given,

i.e. $100a + 10b + c = 6(a + b + c)$

now, we’ll put together the like terms to simplify the

$ \Rightarrow 100a - 6a + 10b - 6b = 6c - c$

$ \Rightarrow 94a + 4b = 5c$

But no value of ‘c’ can satisfy the equation.

Therefore, we get just one solution i.e.,

${\text{ = 10(5) + 4}}$

${\text{ = 50 + 4}}$

${\text{ = 54}}$