Never Prime
If a two digit number has its digits reversed and the smaller of the two numbers is subtracted from the larger, prove the difference can never be prime.
Age
14 to 16
Challenge level
Take any two digit number, reverse its digits, and subtract the smaller number from the larger. For example $$42-24=18$$ I've tried this a few times and I never seem to end up with a prime number. Try some examples of your own. Do you ever end up with a prime number?
Can you prove that you will never end up with a prime?
What happens when I do the same with a three digit number?
What about a four digit number?
What about a five, six, seven, ... $n$ digit number?
Can you justify your findings?
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.