Question

A student of the 5th standard started writing down the counting numbers as 1, 2, 3, 4, .... and then he added all those numbers and got the result 500. But when I checked the result I found that he had missed a number. What is the missing number?

Answer 1

Hint: In order to solve to this problem, we must proceed by the use of formula of sum of first n natural number which is equal to $\dfrac{{{\text{n}}\left( {{\text{n}} + 1} \right)}}{2}$. Since the student did not add one number, the actual sum will be greater than 500.

Complete step-by-step answer:
It is given in the question that 1+2+3+...+n = 500
And we know that
The sum of the first n positive integers
Sum =$\sum\limits_{{\text{r = }}1}^{\text{n}} {{\text{r}} = \dfrac{{{\text{n}}\left( {{\text{n}} + 1} \right)}}{2}}$; where, r varies from 1 to n
${\text{Sum = }}\dfrac{{{\text{n}} \times \left( {{\text{n + 1}}} \right)}}{2}$
We should understand that the sum will be greater than 500, because he left out one number. Therefore,
${\text{Sum = }}\dfrac{{{\text{n}} \times \left( {{\text{n + 1}}} \right)}}{2}{\text{ > 500}}$
Multiply both sides by 2
$\Rightarrow 2 \times \dfrac{{{\text{n}} \times \left( {{\text{n + 1}}} \right)}}{2}{\text{ > 2}} \times {\text{500}}$
$\Rightarrow {\text{n}} \times \left( {{\text{n + 1}}} \right){\text{ > 1000}}$
This can be written as
$\Rightarrow {{\text{n}}^2}{\text{ + n > 1000}}$
$\Rightarrow {{\text{n}}^2}{\text{ + n - 1000 > 0}}$
To find the critical numbers we solve ${{\text{n}}^2}{\text{ + n - 1000 > 0}}$

Here, we have, a quadratic equation ${{\text{n}}^2}{\text{ + n - 1000 = 0}}$
Sridharacharya formula is actually the quadratic formula, used for finding the roots of a quadratic equation ${\text{a}}{{\text{x}}^2}{\text{ + bx + c = 0}}$ , where a not equal to 0 , & a, b, c are real coefficients of the equation ${\text{a}}{{\text{x}}^2}{\text{ + bx + c = 0}}$
Being quadratic it has 2 roots.
X = \dfrac{{\left( { - {\text{b + }}\sqrt {{{\text{b}}^2} - 4{\text{ac}}} } \right)}}{{2{\text{a}}}}{\text{ & }}\dfrac{{\left( { - {\text{b - }}\sqrt {{{\text{b}}^2} - 4{\text{ac}}} } \right)}}{{2{\text{a}}}} (1)
On comparing the given equation ${{\text{n}}^2}{\text{ + n - 1000 = 0}}$ with the general quadratic equation ${\text{a}}{{\text{x}}^2}{\text{ + bx + c = 0}}$ we got values of coefficients a = 1, b = 1, c = -1000
On putting the value of coefficients a, b, c in equation (1)
{\text{n = }}\dfrac{{\left( { - (1){\text{ + }}\sqrt {{{(1)}^2} - 4 \times (1) \times ( - 1000)} } \right)}}{{2 \times 1}}{\text{ & }}\dfrac{{\left( { - (1){\text{ - }}\sqrt {{{(1)}^2} - 4 \times (1) \times ( - 1000)} } \right)}}{{2 \times 1}}
{\text{n = }}\dfrac{{\left( {{\text{ - 1 + }}\sqrt {4001} } \right)}}{2}{\text{ & }}\dfrac{{\left( {{\text{ - 1 - }}\sqrt {4001} } \right)}}{2}
{\text{n = 31}}{\text{.126 & \- 32}}{\text{.126}}
We know that if discriminant D $\geqslant {\text{0}}$ then it will give real and distinct roots.
Here D= ${\text{ = }}\sqrt {{{(1)}^2} - 4 \times 1 \times ( - 1000)} {\text{ = }}\sqrt {4001} {\text{ }} \geqslant {\text{0}}$
Therefore we got two distinct real roots {{\text{n}}_1}{\text{ = 31}}{\text{.126 & }}{{\text{n}}_2}{\text{ = - 32}}{\text{.126}}
We ignore the negative. So the solution is for n > 31.126
So the smallest value of n could be is 32
${\text{S = }}\dfrac{{{\text{n}} \times \left( {{\text{n + 1}}} \right)}}{2}$,
substitute 32 for n,
${\text{S = }}\dfrac{{32 \times \left( {{\text{32 + 1}}} \right)}}{2}$
${\text{S = 528}}$
As it is given in the question that sum is 500 so the student forgot to add 528−500=28

∴ the missing number will be 28

NOTE- This type of particular question can also be solved by other approach which doesn’t include solution of Quadratic equation which states as-
It is given that 1+2+3+...+n=500
⇒${\text{Sum = }}\dfrac{{{\text{n}} \times \left( {{\text{n + 1}}} \right)}}{2}{\text{ = 500}}$
$\Rightarrow {{\text{n}}^2}{\text{ + n = 1000 }}$
$\Rightarrow {\text{ }} \approx {{\text{n}}^2} = 1000$
Since ${31^2} < 1000 < {32^2}$
⇒961<1000<1024
The perfect square closest to 1000 is 1024=${32^2}$

Since the student did not add one number, so 500 is less than the actual sum and 1000 will also be less than the actual value.
Hence, we will take 32
Hence, initially 32 were added.
So their sum will be ${\text{S = }}\dfrac{{32 \times \left( {{\text{32 + 1}}} \right)}}{2}$ ​ = 33×16 = 528
As the given sum is 500 so the student forgot to add 528−500=28
$\therefore$ The missing number is 28