Question

The sum of first $N$ natural numbers is equal to ${x^2}$ where $x$ is an integer less than $100$. What are the values that $N$ can take?
A.$1,9,27$
B. $1,7,26$
C. $1,8,48$
D. $1,8,49$

Answer 1

Hint: We are going to use the formula of the sum of first $N$ natural numbers, that is, $\dfrac{{n(n + 1)}}{2}$.
Here, we cannot find direct values so we will use the hit and trial method to find the solution using the numbers given in the options.

Complete step by step solution :

On observing the question, we understand that the sum of first$N$natural numbers will be equal to${x^2}$where$x < 100$.
We use the aforementioned formula for the sum of first$N$natural numbers,
$\Rightarrow \dfrac{{n(n + 1)}}{2} = {x^2}.......(1)$
We have to use the Hit and Trial method here to find the suitable values of$N$.The numbers that we will use for this method are$1,7,8,9,26,27,48\& 49$.
When $N = 1$,

$${x^2} = \dfrac{{1(1 + 1)}}{2} \\\ \Rightarrow {x^2} = \dfrac{{1 \times 2}}{2} \\\ \Rightarrow {x^2} = 1 \\\ \Rightarrow x = \sqrt 1 = \pm 1 \\\$$

Both the values of $x$are less than 100, therefore $N = 1$.
When $N = 7$,

$${x^2} = \dfrac{{7(7 + 1)}}{2} \\\ \Rightarrow {x^2} = \dfrac{{7 \times 8}}{2} \\\ \Rightarrow {x^2} = 7 \times 4 = 28 \\\ \Rightarrow x = \sqrt {28} = \pm 2\sqrt 7 \\\$$

Both the values of $x$are less than 100, but are not integers. Therefore $N \ne 7$.
When $N = 8$,

$${x^2} = \dfrac{{8(8 + 1)}}{2} \\\ \Rightarrow {x^2} = \dfrac{{8 \times 9}}{2} \\\ \Rightarrow {x^2} = 4 \times 9 = 36 \\\ \Rightarrow x = \sqrt {36} = \pm 6 \\\$$

Both the values of $x$are less than 100, therefore $N = 8$.
When $N = 9$,

$${x^2} = \dfrac{{9(9 + 1)}}{2} \\\ \Rightarrow {x^2} = \dfrac{{9 \times 10}}{2} \\\ \Rightarrow {x^2} = 9 \times 5 = 45 \\\ \Rightarrow x = \sqrt {45} = \pm 3\sqrt 5 \\\$$

Both the values of $x$are less than 100, but are not integers. Therefore $N \ne 9$.
When $N = 26$,

$${x^2} = \dfrac{{26(26 + 1)}}{2} \\\ \Rightarrow {x^2} = \dfrac{{26 \times 27}}{2} \\\ \Rightarrow {x^2} = 13 \times 27 = 351 \\\ \Rightarrow x = \sqrt {351} = \pm 3\sqrt {39} \\\$$

Both the values of $x$are less than 100, but are not integers. Therefore $N \ne 26$.
When $N = 27$,

$${x^2} = \dfrac{{27(27 + 1)}}{2} \\\ \Rightarrow {x^2} = \dfrac{{27 \times 28}}{2} \\\ \Rightarrow {x^2} = 27 \times 14 = 378 \\\ \Rightarrow x = \sqrt {378} = \pm 3\sqrt {42} \\\$$

Both the values of $x$are less than 100, but are not integers. Therefore $N \ne 27$.
When $N = 48$,

$${x^2} = \dfrac{{48(48 + 1)}}{2} \\\ \Rightarrow {x^2} = \dfrac{{48 \times 49}}{2} \\\ \Rightarrow {x^2} = 24 \times 49 = 1176 \\\ \Rightarrow x = \sqrt {1176} = \pm 14\sqrt 6 \\\$$

Both the values of $x$are less than 100, but are not integers. Therefore $N \ne 48$.
When $N = 49$,

$${x^2} = \dfrac{{49(49 + 1)}}{2} \\\ \Rightarrow {x^2} = \dfrac{{49 \times 50}}{2} \\\ \Rightarrow {x^2} = 49 \times 25 = 1225 \\\ \Rightarrow x = \sqrt {1225} = \pm 35 \\\$$

Both the values of $x$are less than 100, therefore $N = 49$.
Therefore, we come to the conclusion that $N = 1,8,49$.
Thus, the answer is option D.
Note: We were given options in this question so we knew for which values we had to check the conditions. But if the options are not given, we will have to use the hit and trial method blindly taking up random numbers. While taking random numbers, we need to remember to check that the conditions are being satisfied and if not, choose numbers which do satisfy the conditions.