Copyright © University of Cambridge. All rights reserved.
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 |
|