Video: AQA GCSE Mathematics Higher Tier Pack 5 β€’ Paper 2 β€’ Question 30

π‘Ž, 𝑏, and 𝑐 are positive, whole numbers. 𝑐 is a prime number. π‘Žπ‘ βˆ’ 𝑏𝑐 is a prime number. Show that π‘Ž and 𝑏 are consecutive numbers.

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Video Transcript

π‘Ž, 𝑏, and 𝑐 are positive whole numbers. 𝑐 is a prime number. π‘Žπ‘ minus 𝑏𝑐 is a prime number. Show that π‘Ž and 𝑏 are consecutive numbers.

First, let’s just be clear on some of the wording in the question. First of all, prime numbers are numbers which have exactly two factors. We use this definition rather than the description that they have factors of one and themselves because this creates some confusion about whether or not one is a prime number. One has factors of one and itself because itself is one. But it doesn’t have exactly two factors. It has only one factor. So one is not a prime number. If we use the definition of a prime number as having exactly two factors, then this doesn’t create any confusion.

However, it is still true that if a number is prime, then its only factors will be one and itself. So if 𝑐 is a prime number, then its only factors will be 𝑐 and one. Equally, if π‘Žπ‘ minus 𝑏𝑐 is a prime number, then its only factors will be itself, π‘Žπ‘ minus 𝑏𝑐, and one. Now, we’re asked to show that π‘Ž and 𝑏 are consecutive numbers. That just means integers that come one after the other, such as six and seven. And to do so, let’s look at this number π‘Žπ‘ minus 𝑏𝑐 in more detail.

We can see that the two terms that we’re subtracting do in fact have a common factor of 𝑐. We can therefore factorize the expression π‘Žπ‘ minus 𝑏𝑐 as 𝑐 multiplied by π‘Ž minus 𝑏. But this would mean that the expression π‘Žπ‘ minus 𝑏𝑐 has factors 𝑐 and π‘Ž minus 𝑏 as well as one and itself. This can’t be the case if π‘Žπ‘ minus 𝑏𝑐 is a prime number. So it must be the case then that 𝑐 and π‘Ž minus 𝑏 are actually equal to one and the original number itself. That’s π‘Žπ‘ minus 𝑏𝑐.

But 𝑐, remember, is prime. So 𝑐 can’t be equal to one as one isn’t a prime number for the reasons we discussed earlier. Therefore, it must be the case that 𝑐 is actually equal to π‘Žπ‘ minus 𝑏𝑐. And the other factor, π‘Ž minus 𝑏, is equal to one. But if π‘Ž minus 𝑏 is equal to one, then adding 𝑏 to each side of this equation, we see that π‘Ž is equal to 𝑏 plus one. But if π‘Ž is equal to 𝑏 plus one, then this means that π‘Ž is the next whole number up after 𝑏, which tells us that π‘Ž and 𝑏 are consecutive. This is what we originally asked to show.

So by factorizing π‘Žπ‘ minus 𝑏𝑐 and then using the fact that both 𝑐 and π‘Žπ‘ minus 𝑏𝑐 were prime, we show that in fact the number 𝑐 is equal to π‘Žπ‘ minus 𝑏𝑐. And π‘Ž minus 𝑏 is equal to one, which means that π‘Ž is equal to 𝑏 plus one, making π‘Ž and 𝑏 consecutive numbers.

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