Find Four Primes Smaller Than 100 Which Are Factors Of 3^32 − 2^32: Maths Olympiad Walkthrough
Maths Olympiad
Maths Olympiads take place worldwide, providing talented young mathematicians with an opportunity to test their skills against different kinds of questions to those they usually face in school.
In this article, we will look at how to solve a challenging number question using just high school / secondary school mathematics.
In this article, we will look at how to solve a challenging number question using just high school / secondary school mathematics.
The Question
Find four prime numbers smaller than 100 which are factors of 33^2 − 23^2.
The Solution: How to begin
With the size of the numbers involved in this question, even if we were allowed to use a calculator, we may find it difficult to find any solutions. 33^2, in particular, is too big a number to use on the vast majority of calculators. It's therefore clear that we need to perform some kind of clever trick to solve this problem.
The first thing to notice about 33^2 − 23^2 is that we have two numbers, both being raised to the same even power before being subtracted one from the other.
This should remind us of a very important skill to be familiar with in high school or secondary mathematics. That is the difference of two squares which works as follows.
(a + b)(a − b) = a^2 + ab − ab − b^2 = a^2 − b^2
This simple equation tells us that if we have one number squared being subtracted from another number squared, we can rewrite them as the sum of the two numbers multiplied by the difference between them. This is going to be very important in solving this problem.
The first thing to notice about 33^2 − 23^2 is that we have two numbers, both being raised to the same even power before being subtracted one from the other.
This should remind us of a very important skill to be familiar with in high school or secondary mathematics. That is the difference of two squares which works as follows.
(a + b)(a − b) = a^2 + ab − ab − b^2 = a^2 − b^2
This simple equation tells us that if we have one number squared being subtracted from another number squared, we can rewrite them as the sum of the two numbers multiplied by the difference between them. This is going to be very important in solving this problem.
Using the difference of two squares
By the laws of indices, if we raise a number to a power and then raise it to another power, the result is the same as if we had raised it to the product of the two separate powers. Algebraically, this looks like this:
(x^a)^b = x^ab = (x^b)^a
We can use this to rewrite our problem:
33^2 − 23^2 = (3^16)^2 − (2^16)^2
We now have a difference of two squares and so:
33^2 − 23^2 = (3^16 + 2^16)(3^16 − 2^16)
We can repeat the trick with our second bracket as again we have the difference of two even powers, hence the difference of the squares of half this power:
33^2 − 23^2 = (3^16 + 2^16)(3^8 + 2^8)(3^8 − 2^8)
We can keep on repeating this; each time splitting the last bracket using our difference of two squares rule. Finally, we get:
33^2 − 23^2 = (3^16 + 2^16)(3^8 + 2^8)(3^4 + 2^4)(3^2 + 2^2)(3 + 2)(3 − 2)
As we now have six brackets multiplying together to give us our original expression, we have six factors. Our next job is to check if any of them are prime and smaller than 100.
3 − 2 = 1 - A trivial factor and not a prime number
3 + 2 = 5 - A prime number
3^2 + 2^2 = 9 + 4 =13 - A prime number
3^4 + 2^4 = 81 + 16 = 97 - A prime number
We now have three prime factors smaller than 100; 5, 13 and 97.
Just one more to go, but to find this we will need to evaluate 3^8 + 2^8.
(x^a)^b = x^ab = (x^b)^a
We can use this to rewrite our problem:
33^2 − 23^2 = (3^16)^2 − (2^16)^2
We now have a difference of two squares and so:
33^2 − 23^2 = (3^16 + 2^16)(3^16 − 2^16)
We can repeat the trick with our second bracket as again we have the difference of two even powers, hence the difference of the squares of half this power:
33^2 − 23^2 = (3^16 + 2^16)(3^8 + 2^8)(3^8 − 2^8)
We can keep on repeating this; each time splitting the last bracket using our difference of two squares rule. Finally, we get:
33^2 − 23^2 = (3^16 + 2^16)(3^8 + 2^8)(3^4 + 2^4)(3^2 + 2^2)(3 + 2)(3 − 2)
As we now have six brackets multiplying together to give us our original expression, we have six factors. Our next job is to check if any of them are prime and smaller than 100.
3 − 2 = 1 - A trivial factor and not a prime number
3 + 2 = 5 - A prime number
3^2 + 2^2 = 9 + 4 =13 - A prime number
3^4 + 2^4 = 81 + 16 = 97 - A prime number
We now have three prime factors smaller than 100; 5, 13 and 97.
Just one more to go, but to find this we will need to evaluate 3^8 + 2^8.
Finding the value of 3^8 + 2^8
The next bracket is trickier as the powers are higher but still answerable without a calculator.
3^8 = 3^4 × 3^4 = 81 × 81
To do this without a calculator, let's think of it as (80 + 1)(80 + 1) and expand brackets.
(80 + 1)(80 + 1) = 80 × 80 + 80 × 1 + 1 × 80 + 1 × 1
= 6400 + 80 + 80 + 1
= 6561
2^8 = 2^4 × 2^4 = 16 × 16 = 256
Therefore 3^8 + 2^8 = 6561 + 256 = 6817
3^8 = 3^4 × 3^4 = 81 × 81
To do this without a calculator, let's think of it as (80 + 1)(80 + 1) and expand brackets.
(80 + 1)(80 + 1) = 80 × 80 + 80 × 1 + 1 × 80 + 1 × 1
= 6400 + 80 + 80 + 1
= 6561
2^8 = 2^4 × 2^4 = 16 × 16 = 256
Therefore 3^8 + 2^8 = 6561 + 256 = 6817
Is 6817 a prime number?
At first glance, it can be difficult to see if 6817 is prime or not, but as our final prime factor needs to be smaller than 100, it seems likely that 6817 can be broken down further into smaller factors.
Upon closer inspection, we can see that both 68 and 17 are multiples of 17 (68 = 4 × 17), and so 6817 itself must also be a multiple of 17 (6817 = 401 × 17, although we don't actually need to know this for our solution). 17 is a prime number smaller than 100, and so we have found our fourth prime factor.
Upon closer inspection, we can see that both 68 and 17 are multiples of 17 (68 = 4 × 17), and so 6817 itself must also be a multiple of 17 (6817 = 401 × 17, although we don't actually need to know this for our solution). 17 is a prime number smaller than 100, and so we have found our fourth prime factor.
The Solution
We have now seen that:
3^32 − 2^32 = (3^16 + 2^16) × 401 × 97 × 17 × 13 × 5
and so the four prime factors we were looking for are 5, 13, 17 and 97.
3^32 − 2^32 = (3^16 + 2^16) × 401 × 97 × 17 × 13 × 5
and so the four prime factors we were looking for are 5, 13, 17 and 97.
Comments
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