Question

How many postive integers less than 100 can be written as the sum of 11 consecutive integers​

Answer 1

$\large\text{\underline{Required Basics}}$

$\red{\bigstar\ \text{The Sum of the First }n\text{ Integers}}$

The sum of the first $n$ positive integers, $\purple{S_{n}}$ is,

$\implies\purple{\dfrac{n(n+1)}{2}}$

$\red{\bigstar\ \text{The Number of Integers Between }x\text{ and }y}$

The number of integers between $x$ and $y$, with $y\geq x$ is,

$\implies\purple{y-x+1}$

$\large\text{\underline{Solution}}$

The sum of the first $n$ integers;

$\implies S_{n}=\dfrac{n(n+1)}{2}$

The sum of the first $n+11$ positive integers;

$\implies S_{n+11}=\dfrac{(n+11)(n+12)}{2}$

The way to obtain the sum of consecutive 11 natural numbers is using the subtraction between the two sums;

$\implies S_{n+11}-S_{n}=\dfrac{n^{2}+23n+132}{2}-\dfrac{n^{2}+n}{2}$

$\implies S_{n+11}-S_{n}=\dfrac{22n+132}{2}$

$\implies S_{n+11}-S_{n}=11n+66$

Hence, this should be less than 100;

$\implies11n+66<100$

On solving,

$\implies 11n<100-66$

$\implies 11n<34$

$\implies n<\dfrac{34}{11}$

We are adding numbers starting from $n+1$;

$\implies n+1\geq1$

$\implies n\geq0$

The common interval of two inequality is,

$\implies 0\leq n<\dfrac{34}{11}$

So, the solution is,

$\implies\boxed{n=0,1,2,3}$

$\large\text{\underline{Conclusion}}$

There are four numbers in the form of $11n+66$, which are $66$, $77$, $88$, or $99$.

$\large\text{\underline{Verification}}$

We are adding numbers from $n+1$ to $n+11$;

$n=0$

$\implies66=1+2+3+4+5+6+7+8+9+10+11$

$n=1$

$\implies77=2+3+4+5+6+7+8+9+10+11+12$

$n=2$

$\implies88=3+4+5+6+7+8+9+10+11+12+13$

$n=3$

$\implies99=4+5+6+7+8+9+10+11+12+13+14$

Hence verified.

Answer 2

Given : positive integers less than 100 can be written as the sum of 11 consecutive integers

To Find : How many

Solution:

from - 5 to + 5 , 11 consecutive integers  sum would be zero

looking for positive integers

Hence

1st case -4 to  + 6    sum = 11

2nd case -3  to 7      sum = 22

3rd case - 2 to 8       sum = 33

4th case -1 to 9        sum = 44

5th case 0 to 10       sum = 55

6th case 1 to 11         sum = 66

7th case 2 to 12       sum = 77    

8th case 3 to 13        sum = 88

9th case 4 to 14          Sum  =  99

9 positive integers  less than 100 can be written as the sum of 11 consecutive integers

if consecutive integers are also need to be positive

Then  4 case

  1 to 11  => Sum = 66

  2 to 12 => Sum = 77

  3 to 13 => sum = 88

 4 to 14 => Sum = 99

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