If the sum of three consecutive integers is 162 how do you find the smallest integer?

Hint: We have given a sum of three consecutive integers so let us assume the smallest integer be n, the next consecutive integer will be (n + 1) and the next integer will be (n + 2). Then add these three integers and equate the addition to 162. After that we will find the value of n. As we have assumed the smallest integer is n then the value of n will give us the value of the smallest integer.

Complete step by step solution:
Let us assume the three consecutive integers as:
$n,n+1,n+2$
From the above integers, the smallest integer is n. As it is given that sum of these three consecutive integers is 162 so adding the above three assumed integers and equating them to 162 we get,
$n+n+1+n+2=162$
Now, adding the three n given in the above equation and the constant term 1 and 2 on the L.H.S of the above equation we get,
$\Rightarrow 3n+3=162$
Subtracting 3 on both the sides we get,
$\begin{align}
  & 3n=162-3 \\
 & \Rightarrow 3n=159 \\
\end{align}$
Dividing 3 on both the sides of the above equation we get,
$\begin{align}
  & \Rightarrow n=\dfrac{159}{3} \\
 & \Rightarrow n=53 \\
\end{align}$
In the above, we have shown that “n” is the smallest integer so the smallest integer which we have calculated above is 53.
Hence, the smallest integer in the above three consecutive integers is 53.

Note: To solve the above problem you must know what the consecutive integers are. Consecutive integers are those integers in which the difference between any two integers is 1. And this difference is calculated by taking any integer and subtracting the previous integer from this integer.
Let us take 4 consecutive integers say,
$n,n+1,n+2,n+3$
Now, let us take an integer n + 1 then the previous integer from n + 1 is n so subtracting n from n +1 we get,
$\begin{align}
  & n+1-n \\
 & =1 \\
\end{align}$