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A number of pleasing solutions were received.
Here is one offered by Andrei of School 205 Bucharest:
I observed that number 234234 is:
234234 = 234x1000 + 234 = 234(1000 + 1) = 1001 x 234
But 1001 is always divisible by 91:
1001 = 7 x 11 x 13 = 77 x 13 = 91 x 11 = 143 x 7
The same is true for 973973:
973973 = 1001 x 973
In fact, if you repeat any 3-digit number in that manner, you obtain a number divisible by 1001, so by 7, 11, 13, 77, 91 and 143:
Now, I look through other patters:
- if you repeat a 2-digit number twice,you obtain a number divisible by 101:
E.g. 65 ...6565
6565/101 = 65
- if you repeat a 2-digit numberthree times you obtain a number divisible by 3, 7, 13, 37 and all combinations obtained by multiplying them (21, 39, 111, 91, 259, 351, 273, and 3367).
- if you repeat a 4-digit number twice
Here, I obtained a prime number.
I did not continue further, as I would obtain too large numbers.
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?
Choose any 3 digits and make a 6 digit number by repeating the 3 digits in the same order (e.g. 594594). Explain why whatever digits you choose the number will always be divisible by 7, 11 and 13.
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.