### Consecutive Numbers

An investigation involving adding and subtracting sets of consecutive numbers. Lots to find out, lots to explore.

### Calendar Capers

Choose any three by three square of dates on a calendar page...

Make a set of numbers that use all the digits from 1 to 9, once and once only. Add them up. The result is divisible by 9. Add each of the digits in the new number. What is their sum? Now try some other possibilities for yourself!

# Kids

##### Age 11 to 14Challenge Level

There were lots of good solutions to this question. You have to replace the seven letters by seven numbers.

 M U M + D A D -- -- -- -- K I D S

Congratulations to Martina Murtagh, age 14, Our Lady's School, Newry who found all 12 solutions (and a few more taking K=0). Bei Guo, age 14 from Riccarton High School, Christchurch, New Zealand sent an excellent solution. Bei noted that the solutions come in 6 pairs because you can keep all the other numbers the same and exchange the values of the middle letters A and U. George Vassilev, year 6, Rosebank Primary School, Leeds and students from the River Valley High School, Singapore and from Bourne Grammar School found some of the solutions. The following well explained write-up is by Michael Brooker, age 9 (home-educated) who found all 12 solutions.

M + D = S
U + A + any remainder = D
M + D + any remainder = KI

For I to be different from S, there must be a remainder. If S = 9 and KI = 10 then M + D must obviously be 9, not 19.

So the sum is at the moment looking like this:

 4 ? 4 + 5 ? 5 -- -- -- -- 1 0 5 9

or something fairly similar.

These ones do not work because they leave two letters representing the same number:

 1 ? 1 + 8 ? 8 -- -- -- -- 1 0 8 9
 8 ? 8 + 1 ? 1 -- -- -- -- 1 0 1 9
 0 ? 0 + 9 ? 9 -- -- -- -- 1 0 9 9
 9 ? 9 + 0 ? 0 -- -- -- -- 1 0 0 9

These do not work because to get a remainder, U or A would have to have the same value as S:

 2 ? 2 + 7 ? 7 -- -- -- -- 1 0 7 9
 3 ? 3 + 6 ? 6 -- -- -- -- 1 0 6 9

This leads to eight possible solutions:

 4 8 4 + 5 7 5 -- -- -- -- 1 0 5 9
 4 7 4 + 5 8 5 -- -- -- -- 1 0 5 9
 5 8 5 + 4 6 4 -- -- -- -- 1 0 4 9
 5 6 5 + 4 8 4 -- -- -- -- 1 0 4 9
 6 8 6 + 3 5 3 -- -- -- -- 1 0 3 9
 6 5 6 + 3 8 3 -- -- -- -- 1 0 3 9
 7 8 7 + 2 4 2 -- -- -- -- 1 0 2 9
 7 4 7 + 2 8 2 -- -- -- -- 1 0 2 9

I found four more solutions by choosing another number that M and D could add up to. They obviously can't add up to 10 or 11, because that would leave two letters representing the same number. For the same reason, they also can't add up to 13, 14, 15, 16, 17 or 18. They can't add up to 19 for a different reason: there is no such thing as a digit greater than nine (except in hexadecimal).

Therefore the only other number that M and D can add up to is 12. I took a closer look and found that there were only two pairs of numbers which add up to 12 without leaving two letters representing the same number. However, there are four solutions with 12 because each pair of digits provides two solutions.

 7 6 7 + 5 8 5 -- -- -- -- 1 3 5 2
 7 8 7 + 5 6 5 -- -- -- -- 1 3 5 2
 8 6 8 + 4 7 4 -- -- -- -- 1 3 4 2
 8 7 8 + 4 6 4 -- -- -- -- 1 3 4 2