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.
What is the sum of all the digits in all the integers from one to
The following solution was sent in by Julia of
Downe House School. Well done Julia!
The total reached when the complete cycle of numbers are added
together must always be the same because:
At one point in the cycle, each number (1 to 9) will be in each
and every digit position ( e.g., units, tens, hundreds,
thousands etc. ). In the first position for example,
whatever number is there in the first place, by the end of the
cycle all the numbers from 1 to 9 will have been there and so the
total in that column will always be the same (the sum of the
numbers 1 to 9, i.e. 45). The same is true for each column or digit
position so the overall total will always be the same. The units
digit is 5, then 4 is carried into the tens place to make 49, so
the tens digit is 9 and 4 is carried into the hundreds place ...
and so on.
The total is 4 999 999 995.