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## Pre-algebra

### Course: Pre-algebra>Unit 12

Lesson 4: Equations word problems

# Sum of integers challenge

Sal solves the following problem: The sum of three consecutive odd integers is 231. What is the largest integer? Created by Sal Khan and Monterey Institute for Technology and Education.

## Want to join the conversation?

• I think there is a little mistake.
The statement says : the sum of three consecutive odd integers is 231. But Sal is using even integers which is x+2 and x+4 to solve for this question.
• x, x+2, and x+4 are just any 3 consecutive numbers that differ by 2. They're neither even nor odd.
It's the fact that they add up to 231 that makes them odd - if they added up to 234 they would be even.
• what is the sum off all integers from 1 to 100
• hi i have a question.

find three consecutive odd numbers such that the sum of 3/5 of the first, 1/2 of the second and 3/8 of the third is 63.
i tried this and got so many different weird answers
• Let us denote our variables as a, a+2, and a+4.
Based on your equation, we have 3/5a + 1/2(a+2) + 3/8(a+4) = 63
Simplifying, we get 3/5a + 1/2a + 1 + 3/8a + 3/2 = 63.
Then get a common denominator:
24/40a + 20/40a + 1 + 15/40a + 60/40 = 63
Then, adding, 59/40a + 1 + 3/2 = 63
Subtracting the integers, 59/40a = 121/2
Multiplying leaves us with 59a = 2420
Finally, dividing, a = ~41.02.
Therefore, there are no integers that will satisfy this equation, only repeating decimals.
Hope this helps
• Can someone please solve this equation that I found online. It's hard.

*The product of 3 whole numbers is 72. What is the maximum possible sum of these three whole numbers.*
• In how many ways can you express 72 as the product of 3 natural numbers (unordered ... Then the possible cases are ... How do you find the sum of all natural numbers amongst first one thousand ... The number of ways the 3 powers of 2 is distributed over the three variables ... What is the maximum value of their product?
• Hi. I’m struggling with how the problem below is solved. I have the answer but not sure how it was worked out. I’ve only worked problems where the sum was given. I want to understand how they got the answer.

Problem: Two times the sum of three consecutive odd integers is the same as 23 more than 5 times the largest integer. Find the integers.
• Although I did not calculate the equation, I think you can write the equation like this:
2(x+x+2+x+4)=5(x+4)
Hope it gives a sense of how to solve the problem!
(1 vote)
• how'd you get +6? in the equation?
• He had the equation x + x + 2 + x + 4 = 231 , then he added the Xs together to get 3x + 2 + 4 = 231, then the integers:
3x + 6 = 231
• two consecutive odd integers sum to -8 what are they? guys can I have an answer plz
• Odd integers increase by 2. So you need to use X=1st odd integer and X+2=2nd odd integer
"Sum" means you add them to get -8
X + X + 2 = -8
Now, solve for X to find the 1st number.
Then, add 2 to find the 2nd number.
Hope this helps.
• I have a weird way of doing it. Not sure if it is correct? Pls tell me if it is not right. So for Sal's question there are 3 odd consecutive integers and the sum 231. so I divided 231 by 3 and got 77. then I go on and do 77+79+81=237 which is not equivalent to 231. I move a number back 75+77+79=321 and I get my answer
• Your method works. The answer you find would be the middle number. You are basically treating the starting number as the total to find an average. Dividing 3 finds the average result (which will be the middle number).

The potential issue is that is you were instructed to find the answer algebraically, then you aren't. Solving algebraically would require finding the pattern in the numbers and using a variable to create an algebraic equation, then solve that equation. It pays to learn / understand how to do this because there are other word problems that you must solve algebraically (not direct mathematical technique).
• At , I would have added 6 to both sides of the equation rather than subtracted 6 because that would get 3x + 12 on the left-hand side. Then, you can divide both sides by twelve to get x + 4, which is the largest integer.
Also, I may have set up the original equation as x+(x-2)+(x-4)=231, because then x= the largest integer and it will be easier to answer the question. Do you agree with my methods? Thanks!
• Yes. Very clever but even easier is divide by three to get middle one.
(1 vote)
• I thought that the number 1 is an odd number because in both videos you use 3 as the smallest odd iteger instead of 1. Why is that so?
• When Sal says that 𝑥 is the smallest odd integer, he means that it is the smallest of the three consecutive odd integers that add to 231.

Then to find the other 2 odd integers, he uses 𝑥 = 3 as an example.
If the smallest of three consecutive odd numbers is 3,
then the others are 5 and 7, which we can write as 3 + 2 and 3 + 4.

Thus, if 𝑥 is the smallest of three consecutive odd integers,
then the other two are 𝑥 + 2 and 𝑥 + 4.

– – –

By the way, 1 is only the smallest positive odd integer.
−1 is also an odd integer, as are −3, −5, −7 and so on, forever.
So, in fact, there is no smallest odd integer.