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### Course: Equations & Inequalities 229-236>Unit 2

Lesson 5: Equations word problems

# Sums of consecutive integers

When solving a math problems involving consecutive integers, we can use algebra to set up an equation to solve for the smallest of the four integers. Created by Sal Khan.

## Want to join the conversation?

• a man drove 48 more miles an hour than the number of hours he drove. if he drove 208 miles in all, how many hours did he drive?what was his rate?
(34 votes)
• 48/1=208/?..... Approximately 4 hours and 20 minutes, I guess. His rate was 48 miles per hour and 4.33333 hours per 208 miles.
(22 votes)
• Is it not more simple to divide 136 by 4 and take the two odd pairs on either side of the result?
(31 votes)
• It could be, but he did it via the counting method.
(16 votes)
• I used a different method... The results are the same, but the method itself is a bit more complex. We can assume that an even integer comes in the form of 2n and hence an odd number is 2n+1. Using this method, the four integers are:
2n+1
2n+3
2n+5
2n+7
Their sum is 8n+16=136.
8n=120
n=15
And then the first number is 2n+1=2*15+1=30+1=31
the second number: 2n+3=31+2=33
and so on.

The reason I'm putting this in the questions section is because I was wondering if there is a moment when this method is better than the one presented... Or is it just a more complicated way of doing something simple?
(20 votes)
• In math there are often many different approaches to solving a given problem. Your approach is fine and if it seems more intuitive for you, go for it. Others may or may not find your way more intuitive than what is demonstrated in the video, so thanks for sharing your solution!
(25 votes)
• how do you know what the next integer is going to be like if its x+(x+1)+(x+2) or if its x+(x+2)+(x+4)
(12 votes)
• If it's [consecutive integers], it's going to be x+(x+1)+(x+2), since the common difference between consecutive integers is 1.
If it's [consecutive odd/even integers], it's going to be x+(x+2)+(x+4), since the common difference between odd or even integers is 2.
(26 votes)
• When Khan requested that I pause and attempt to solve the problem on my own, here's what I did:

1) Divided 136 by 4 (ie took the average of the four consecutive odd ints)
2) Result: 34
3) Formed a mental number line around 34 like so: 31-32-33-34-35-36-37
4) Selected the two nearest odd ints to the left of 34 and the two nearest odd ints to the right on my mental number line.
5) added them up to confirm they equal 136

Is there anything wrong with this approach? Are their any scenarios where this approach doesn't work for INTEGERS? (obv. fractions are another matter and Sal's algebraic approach is preferred then).
(11 votes)
• Yes, your method will always work for integers -- and if you can picture the number line in your head clearly enough then often it will be quicker, too. The algebraic method is very useful, though, with situations where it's hard to picture the problem in your head.
(7 votes)
• this video makes no sense
(12 votes)
• Watching it twice and reviewing previous lessons can help with understanding. :)
(3 votes)
• So, when doing consecutive evens we skip count evens x + x+2, x+4, x+6, ....etc? Then odds we use x, x+1, x+2, x+3? I noticed the set up was different depending on what the problem asked (evens/odds)? How can we tell the difference?
(8 votes)
• no, odds are still x, x+2, x+4 etc. also because odd numbers are two apart from each other just like even numbers are. if x =3, then x+1 = 4, but that is not odd. The second pattern you used is for consecutive numbers only.
(7 votes)
• What integer do you start with?
(6 votes)
• In these problems, you don't know the value of the integers. Since consecutive integers are 1 unit higher from each other, you leverage that pattern to define variable representations for the unknown integers:
x = 1st integer
x+1 = 2nd integer
x+2 = 3rd integer
etc.

If the problem deals with consecutive odd or even integers, these increment by 2. So the pattern becomes:
x = 1st odd/even integer
x+2 = 2nd odd/even integer
x+4 = 3rd odd/even integer
etc.

You can extend the pattern to have as many integers as you need for the problem you are doing. Sal needed to find 4 odd consecutive integers, so he extended the pattern out to x+6. Then, apply the math described in the word problem to set up an equation. Sal's problem, asked for the sum of the numbers, so his equation became adding the 4 unknown consecutive odd integers.

Hope this helps.
(9 votes)
• The sum of three consecutive integers is 53 more than the least of the integers. How would I do this?
(4 votes)
• So, you know you have the equation:
x+y+z=x+53. Since x, y, and z are consecutive you also know y=x+1 and z=x+2, so it then becomes x+(x+1)+(x+2), combine to get 3x+3=x+53, subtract x and 3 from both sides to get 2x=50, and x=25, y=26, z=27.
Hope this helps
(8 votes)
• Is it not more simple to divide 136 by 4 and take the two odd pairs on either side of the result?
(5 votes)
• Yes, sometimes algebraic word problems can be solved without algebra and using variables. However, the purpose of the lesson is for you to learn now to solve the problem algebraically. This requires that you learn the skill to find patterns and create an algebraic equation to solve. This helps you to develop basic problem solving skills that you will need when you do algebraic word problems that must be solved using algebra.
(6 votes)

## Video transcript

The sum of four consecutive odd integers is 136. What are the four integers? So before even attempting to tackle it, let's think about what it means to be a consecutive odd integer, what four consecutive odd integers could be referring to. So we could start, let's say, we had 3. The next odd integer-- so we want consecutive odd integers. So we want the next odd integer-- would be 5. Then the one after that would be 7, and the one after that would be 9. These are consecutive odd integers. Another example-- we could start at 11. Then the next odd integer is 13. The next one is 15. The next one is 17. The example of non-consecutive odd integers, if someone went from 3 straight to 7, these are not consecutive. The next odd integer after 3 is 5, not 7. So these are examples of consecutive odd integers. These are not consecutive odd integers. So with that out of the way, let's actually try to tackle this question. And I encourage you to pause this video right now and try to tackle this on your own before I take a go at it. Well as you can imagine, a little bit of algebra might be handy here. So let's let x be equal to the smallest of the four. So if x is the smallest of the four consecutive odd integers, how can we express the other three in terms of x? We'll just use these examples right over here. If that is x, then how would we express this in terms of x? Well, this would be x plus 2. The next odd integer is going to be the previous odd integer plus 2. You're essentially skipping the next number, which is going to be even. So if you added 1, you'd just get to an even number, so you have to add 2 to get to the next odd one. Well if you add 2 again, if you add 2 to x plus 2, well now you get to x plus 4. You add 2 to that. You get to x plus 6. And you see that again right over here. If this was x, then this is x plus 2. This is x plus 4, and this is x plus 6. So in general, if x is the smallest of the integers, we can define the other three as x plus 2, x plus 4, and x plus 6. And let's take the sum of them and set that equal to 136 and then solve for x. So the smallest one is x. Now the next smallest one is going to be x plus 2. Now the one after that's going to be x plus 4, and then the one after that is going to be x plus 6. So this is the sum of the four consecutive odd integers, and they tell us that that is going to be equal to 136. This is equal to 136. And now we can just literally solve for x. We have this one unknown here. So let's add these x terms together. So we have one x, two x's, three x's, four x's. So we can rewrite those literally as 4x. And then we have 2 plus 4, which is 6, plus another 6 is 12. 4x plus 12 is equal to 136. So to solve for x, a good starting point would be to just to isolate the x terms on one side of the equation or try to get rid of this 12. Well to get rid of that 12, we'd want to subtract 12 from the left-hand side. But we can't just do it from the left-hand side. Then this equality wouldn't hold anymore. If these two things were equal before subtracting the 12, well then if we want to keep them equal, if we want the left and the right to stay equal, we've got to subtract 12 from both sides. So subtracting 12 from both sides gives us, well on the left-hand side, we're just left with 4x. And on the right-hand side, we are left with 136 minus 12 is 124. Did I do that right? Yeah, 124. So what is x? Well, we just divide both sides by 4 to solve for x. And we get-- do that in the same, original color-- x is equal to 124 divided by 4. So let's see. 100 divided by 4 is 25. 24 divided by 4 is 6. 25 plus 6 is 31. And if you don't feel like doing that in your head, you could also, of course, do traditional long division. Goes into 124-- 4 doesn't go into 1. 4 goes into 12 three times. 3 times 4 is 12. You subtract, bring down the next 4. 4 goes into 4 one time. You get no remainder. So x is equal to 31. So x is the smallest of the four integers. So this right over here, x is 31. x plus 2 is going to be 33. x plus 4 is going to be 35. And x plus 6 is going to be 37. So our four consecutive odd integers are 31, 33, 35, and 37.