# Video: AQA GCSE Mathematics Foundation Tier Pack 1 β’ Paper 2 β’ Question 11

Three positive integers π, π, and π, sum to 80. π is a square number. π is four times larger than π. Find the values of π, π, and π.

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### Video Transcript

Three positive integers π, π, and π sum to 80. π is a square number. π is four times larger than π. Find the values of π, π, and π.

Letβs begin by formulating some equations based on the information given in the question. Weβre told first of all that three positive integers π, π, and π sum to 80. We can express this as π plus π plus π is equal to 80.

Next, weβre told that π is a square number. Remember a square number is the result of multiplying an integer by itself. We canβ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. Weβ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 canβt be the case that π is bigger than 80. The next statement tells us that π is four times larger than π. We can express this as π is equal to four π. Thatβs four times π.

Now, letβs look at these two equations first of all. What we can do is substitute four π for π in equation one so that we eliminate one of the letters. π plus π plus π is equal to 80. And swapping π for four π gives π plus four π plus π is equal to 80. Four π plus π can be simplified to five π. So we have that π plus five π is equal to 80. And if we subtract π from each side of this equation, we have that five π is equal to 80 minus π.

Now how does this help us? Well, firstly, notice that five π will always be a multiple of five because weβre taking an integer and multiplying it by five. On the other side of the equation, we have 80 minus π. And remember π 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 doesnβ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 π is equal to 80 minus π. And now, we know that 55 is equal to 80 minus 25. On the left, this tells us that five π is equal to 55. And on the right, we see that π is equal to 25.

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

Letβs perform a quick check that these two values of π, π, and π do indeed satisfy all the given conditions β theyβre all positive integers. If we add them together, we do indeed get 80. π is equal to 25 and 25 is equal to five squared. So π is indeed a square number. π is indeed four times larger than π because we use the equation π equals four π to calculate the value of π from the value of π.

So the values of π, π, and π which satisfy all of the required conditions are π equals 25, π equals 44, π equals 11.