# Video: AQA GCSE Mathematics Foundation Tier Pack 1 • Paper 2 • Question 11

Three positive integers 𝑎, 𝑏, and 𝑐, sum to 80. 𝑎 is a square number. 𝑏 is four times larger than 𝑐. Find the values of 𝑎, 𝑏, and 𝑐.

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### Video Transcript

Three positive integers 𝑎, 𝑏, and 𝑐 sum to 80. 𝑎 is a square number. 𝑏 is four times larger than 𝑐. Find the values of 𝑎, 𝑏, and 𝑐.

Let’s begin by formulating some equations based on the information given in the question. We’re told first of all that three positive integers 𝑎, 𝑏, and 𝑐 sum to 80. We can express this as 𝑎 plus 𝑏 plus 𝑐 is equal to 80.

Next, we’re told that 𝑎 is a square number. Remember a square number is the result of multiplying an integer by itself. We can’t directly form this as an equation. But we can write down the first few square numbers to help: one, four, nine, 16, 25, 36, 49, and 64. We’ll stop at this point because the next square number is 81, which is bigger than 80. And as the three numbers are positive and sum to 80, it can’t be the case that 𝑎 is bigger than 80. The next statement tells us that 𝑏 is four times larger than 𝑐. We can express this as 𝑏 is equal to four 𝑐. That’s four times 𝑐.

Now, let’s look at these two equations first of all. What we can do is substitute four 𝑐 for 𝑏 in equation one so that we eliminate one of the letters. 𝑎 plus 𝑏 plus 𝑐 is equal to 80. And swapping 𝑏 for four 𝑐 gives 𝑎 plus four 𝑐 plus 𝑐 is equal to 80. Four 𝑐 plus 𝑐 can be simplified to five 𝑐. So we have that 𝑎 plus five 𝑐 is equal to 80. And if we subtract 𝑎 from each side of this equation, we have that five 𝑐 is equal to 80 minus 𝑎.

Now how does this help us? Well, firstly, notice that five 𝑐 will always be a multiple of five because we’re taking an integer and multiplying it by five. On the other side of the equation, we have 80 minus 𝑎. And remember 𝑎 is a square number. So the right-hand side of the equation is 80 minus a square number.

What we can do now is take that list of square numbers that we wrote down earlier on, subtract each one from 80, and see if we get a multiple of five as the result. Our first square number is one. And 80 minus one gives 79 which is not a multiple of five as it doesn’t end in either five or zero.

Our second square number is four. 80 minus four is equal to 76 which is also not a multiple of five. Our next square number is nine. 80 minus nine is equal to 71, still not a multiple of five. Then, we have 80 minus 16 which gives 64, again not a multiple of five. The next square number in our list is 25. Subtracting this from 80 gives 55, which is a multiple of five. 55 is equal to five times 11.

We can perform a quick check that if we subtract the other square numbers less than 80 from 80, none of them give an answer which is a multiple of five. What does this tell us then? Well, it tells us two things. Well, remember we were solving the equation five 𝑐 is equal to 80 minus 𝑎. And now, we know that 55 is equal to 80 minus 25. On the left, this tells us that five 𝑐 is equal to 55. And on the right, we see that 𝑎 is equal to 25.

To find the value of 𝑐, we need to divide our equation on the left by five. Five 𝑐 divided by five gives 𝑐 and 55 divided by five gives 11. So now, we have two of our values: 𝑎 is equal to 25 and 𝑐 is equal to 11. To find the value of 𝑏, we can return to our equation 𝑏 is equal to four 𝑐. 𝑏 is, therefore, equal to four multiplied by 11 which is equal to 44.

Let’s perform a quick check that these two values of 𝑎, 𝑏, and 𝑐 do indeed satisfy all the given conditions — they’re all positive integers. If we add them together, we do indeed get 80. 𝑎 is equal to 25 and 25 is equal to five squared. So 𝑎 is indeed a square number. 𝑏 is indeed four times larger than 𝑐 because we use the equation 𝑏 equals four 𝑐 to calculate the value of 𝑏 from the value of 𝑐.

So the values of 𝑎, 𝑏, and 𝑐 which satisfy all of the required conditions are 𝑎 equals 25, 𝑏 equals 44, 𝑐 equals 11.