### Repeaters

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.

### Digit Sum

What is the sum of all the digits in all the integers from one to one million?

# Football Sum

##### Stage: 3 Challenge Level:

FOOT + BALL = GAME

We had many solutions to this problem worked out using the fact that the letter O has to be zero or nine.

Also if O=0 then $M=L+1$. Therefore $T+L > 11$ implying that $L \leq8$ and $T \geq 3$ . Therefore the only possible $T, L$ pairs are (3,8) (4,7) (4,8) (5,7) (5,8) (6,7) (6,8) (7,4) (7,5) (7,8) (8,3) (8,4) (8,5) (8,6) (9,3) (9,4) (9,5) (9,6) (9,7).

If O=9 then $M=L-1$. Therefore $T+L \leq 8$. We find that the only possible $T, L$ pairs are (1,6) (2,4) (2,5) (3,1) (4,1) (4,3) (5,1) (5,2) (5,3) (6,1) (7,1).

Each member of the class could take on a different T, L pair and you could all pool your solutions.

Systematically checking all these cases, having found O, T, L and M, will easily give all the solutions and we will know there can't be any others. There are 224 different solutions.

A special mention must go to Helen from Maidstone Girls' Grammar School for her very clear explanation of her use of arithmetic and logic to get answers. Michael of Aranmore Catholic Primary School, Western Australia pointed out that there are many solutions to this problem and sent the examples listed below.

1009   2009   1009   7009   1009   5009   1009   6009

2466   1466   7244   1244   5833   1833   6533   1533

----   ----   ----   ----   ----   ----   ----   ----

3475   3475   8253   8253   6842   6842   7542   7542

1009   2009   1009   7009   1009   5009   1007   2007

2577   1577   7355   1355   5733   1733   2688   1688

----   ----   ----   ----   ----   ----   ----   ----

3586   3586   8364   8364   6742   6742   3695   3695

1009   7009   1007   3007   2990   3990   2009   6009

7466   1466   3688   1688   3155   2155   6144   2144

----   ----   ----   ----   ----   ----   ----   ----

8425   8425   4695   4695   6148   6148   8153   8153

1995   4995   1995   4995   1006   4006   8009   1009

4822   1822   4733   1733   4277   1277   1244   8244

----   ----   ----   ----   ----   ----   ----   ----

6817   6817   6728   6728   5283   5283   9253   9253

3008   6008   2991   3991   2991   3991   2008   3008

6144   3144   3488   2488   3577   2577   3966   2966

----   ----   ----   ----   ----   ----   ----   ----

9152   9152   6479   6479   6568   6568   5974   5974

2004   3004   2003   4003   3911   4991   3995   4995

3677   2677   4788   2788   4066   3066   4211   3211

----   ----   ----   ----   ----   ----   ----   ----

5681   5681   6791   6791   8057   8057   8206   8206
etc

Alexander of Shevah-Mofet'', Tel Aviv also pointed out that
there are many solutions and sent us a more interesting problem to
which he claims there is only one solution. You might like to try
it:


 DONALD + GERALD ROBERT