Or search by topic
Choose any 3 digits and make a 6 digit number by repeating the 3 digits in the same order (e.g. 594594).
Whatever digits you choose the number will always be divisible by 7 and by 11 and by 13, in each case without a remainder.
Can you explain why?
Choose two digits and arrange them to make two double-digit numbers. Now add your double-digit numbers. Now add your single digit numbers. Divide your double-digit answer by your single-digit answer. Try lots of examples. What happens? Can you explain it?
Three people chose this as a favourite problem. It is the sort of problem that needs thinking time - but once the connection is made it gives access to many similar ideas.
Take any four digit number. Move the first digit to the end and move the rest along. Now add your two numbers. Did you get a multiple of 11?