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The sum of eleven consecutive integers is 11. What is the least of these eleven integers?

ant101 May 16, 2018

#1**+2 **

We have:

\(x-5+x-4+x-3+x-2+x-1+0+x+1+x+2+x+3+x+4+x+5\)

Every number cancels out, so we are left with 11x.

So, 11x=11

x=1

Thus, the least of the eleven integers is: \(1-5=\boxed{-4}\)

.tertre May 16, 2018

#2**+2 **

Hi ant101!

The sum of eleven consecutive integers is 11. What is the least of these eleven integers?

Since there are 11 consecutive integers, we can assign a variable to one of the numbers, then write all the other numbers in terms on the variable.

Since there are an odd number of integers, we can call the middle one n.

The five numbers greater than n are in order from least to greatest: \(n+1, n+2, n+3, n+4, n+5\)

Also, the five numbers less than n are in order from least to greatest: \(n-5, n-4, n-3, n-2, n-1\)

Adding them up, we have:

\((n-5)+(n-4)+(n-3)+(n-2)+(n-1)+n+(n+1)+(n+2)+(n+3)+(n+4)+(n+5)=11n.\)

Therefore the sum of the eleven integers can be expressed as 11n.

As you said, the sum of these integers is 11, so we have:

\(11n=11\\ n=1\)

The least of these integers is \(n-5, \text{so} \ 1-5=\boxed{-4}\)

I hope this helped,

Gavin

GYanggg May 16, 2018