#### You may also like ### Upsetting Pitagoras

Find the smallest integer solution to the equation 1/x^2 + 1/y^2 = 1/z^2 ### Rudolff's Problem

A group of 20 people pay a total of £20 to see an exhibition. The admission price is £3 for men, £2 for women and 50p for children. How many men, women and children are there in the group? ### Euler's Squares

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

# Never Prime

##### Age 14 to 16Challenge Level

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.