Question

If the sum of three consecutive integers is 63. Find the smallest number.
(a) 11
(b) 15
(c) 23
(d) 20

Answer 1

Hint: We are asked to look for the smallest number. We will assume our smallest number is x, then we will find its consecutive number as consecutive numbers are the one which differs by 1. So, we get the second and the third number as x + 1 and x + 2 respectively. At last, we add x, x + 1 and x + 2 and equate with 63 to get the value of x.

Complete step-by-step solution:
We are given the sum of three consecutive numbers is 63. We are asked to find the smallest of them. Now, first, we will define the consecutive numbers. The numbers which are just near to each other are called consecutive numbers, they differ by 1 from each other.
Let us start with the assumption that the smallest integer is x. Now the second number is the number consecutive to it. As we know that the consecutive number just differ by 1, so the number next to x will be x + 1, so our second number is x + 1.
Now for the third number, we have to find the consecutive number of x + 1. Again the consecutive number differ by 1, so we just need to move one ahead of x + 1.
So, we get the third number as x + 1 + 1 that is x + 2. So, we have our required number as x, x + 1 and x + 2. We are given that their sum is 63 which means,
\[x+x+1+x+2=63\]
Simplifying further, we get,
\[\Rightarrow 3x+3=63\]
Solving for x, we get,
\[\Rightarrow 3x=63-3\]
\[\Rightarrow 3x=60\]
\[\Rightarrow x=20\]
So, we get x as 20. Hence the smallest number is 20.
Therefore, the correct option is (d).

Note: Remember that the consecutive numbers differ by 1, so do not assume x, 2x, 3x as consecutive as they do not differ by 1, they differ by x. ‘Differ by’ means that the difference between the number is for example, in x, 2x, 3x, 2x – x = x, the difference is x which is not 1.