Question

The sum of three consecutive even numbers is 48. What is the smallest of these numbers?

Answer 1

Hint: Here in this question, we have to find the three numbers which sum is equal to 48. To solve this, in general take any three even numbers i.e., \[x\], \[x + 2\] and \[x + 4\] add these three numbers which equal to 48, simplify using basic arithmetic operation to get the value of \[x\] which is the smallest among the three even number.

Complete step-by-step answer:
Even number is any number that gives a remainder of zero when divided by 2.
Examples: 0, 2, 4, 8, …
Let us take \[x\] to be any even number which can be divided by 2 leaves a remainder 0.
The next two consecutive even numbers after \[x\] are \[x + 2\] and \[x + 4\].
Consider \[x\], \[x + 2\] and \[x + 4\] are three consecutive number given whose sum is equal to 48 i.e.,
\[ \Rightarrow \,\,x + \left( {x + 2} \right) + \left( {x + 4} \right) = 48\]
\[ \Rightarrow \,\,x + x + 2 + x + 4 = 48\]
On simplification, we have
\[ \Rightarrow \,\,3x + 6 = 48\]
Subtract both side by 6, then
\[ \Rightarrow \,\,3x = 48 - 6\]
\[ \Rightarrow \,\,3x = 42\]
Divide both side by 3, we get
\[ \Rightarrow \,\,x = \dfrac{{42}}{3}\]
\[ \Rightarrow \,\,x = 14\]
Hence, the three consecutive numbers are:
\[x = 14\]
\[x + 2 = 14 + 2 = 16\]
\[x + 4 = 14 + 4 = 18\].
Therefore, among three consecutive even numbers 14, 16 and 18 the smallest even number is 14.
So, the correct answer is “14”.

Note: Consecutive integers are integers that follow in sequence, each number being 1 more than the previous number, represented by n, n + 1, n + 2, n + 3, ..., where n is any integer. In similar way
If we start with an even number and each number in the sequence is 2 more than the previous number then we will get consecutive even integers.
If we start with an odd number and each number in the sequence is 2 more than the previous number then we will get consecutive odd integers.