You may also like

problem icon

Euler's Squares

Euler found four whole numbers such that the sum of any two of the numbers is a perfect square...

problem icon

Diophantine N-tuples

Can you explain why a sequence of operations always gives you perfect squares?

problem icon

There's a Limit

Explore the continued fraction: 2+3/(2+3/(2+3/2+...)) What do you notice when successive terms are taken? What happens to the terms if the fraction goes on indefinitely?

Never Prime

Stage: 4 Challenge Level: Challenge Level:2 Challenge Level:2

Good solutions here came from Sam, Dave, Richard and Joe, Matthew and Ross, all at Madras College, St Andrews. Other people tested special cases but did not prove the general results.

If a 2 digit number has its digits reversed and the smaller of the two numbers is subtracted from the larger we prove that this difference can never be prime.

Let the 2 digit number be $a$ where $a> b$. Then $$ab - ba = (10a + b) - (10b + a) = 9(a - b).$$

As $9(a - b)$ is a multiple of $9$, it is not prime.

Now let the 3 digit number be $abc$ $$abc - cba = (100a + 10b + c) - (100c + 10b + a) = 99 (a - c).$$ As $99(a - c)$ is a multiple of $99$, it is not prime.

The 4 digit number can be taken as $abcd$. $$abcd - dcba = (1000a + 100b + 10c + d) - (1000d + 100c + 10b + a) = 9(111a + 10b - 10c - 111d).$$ Again, for any 4 digit number, this difference is a multiple of 9 and so it can't be a prime number.

Similarly for 5 digit numbers: $$\eqalign { abcde - edcba &= (10,000a + 1000b + 100c +10d + e) - (10,000e + 1000d + 100c + 10b + a) \cr &= 99(101a + 10b - 10d - 101e).}$$

This number is a multiple of 99 so it will never be prime.