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Aba

Age 11 to 14 Challenge Level:

It was good to see so many attempts at the solution to this problem. Try to remember that it is important to explain to the reader each of the steps you take. Whilst many of you used some element of trial and improvement in your method, which is fine, you did not explain some of the things that you worked out before you started trying different values. For example, many of you knew that E = 1 and it was worth explaining why this was the case. In the same way you were able to state that B=9 but only a few of you tried to tell me how you knew this. Look at the detail in the solution below, it may help you next time.

Finally, if several of you are sending in solutions from a single school it might be worth looking at what each of you has done to see if there are any patterns that you could use to extend your answers and then send in a joint solution. Often the result will be greater than each of the parts and it is good to talk about what you are doing.

Congratulations to:
Clement Goh of the River Valley School, Singapore;
Elisabeth Elvidge, Alice Lewin and Alice Unwin of the Mount School, York;
Danielle Yule, Laura Lancaster, Briony Pollard and Emma Dellany of the Mount School, York;
Prateek Mehrotra of Riccarton High School, New Zealand;
Andrei Lazanu of School number 205, Bucharest;
for identifying all the possible values for the digits. I have used Andrei's, Prateek's and Clement's solutions as the basis of what is below.

Solution We know that E =1 because this is the greatest amount that can be carried in an addition sum with two numbers..

From the units:

B + D = F (1)
or B + D = 10 + F (2)

From the hundreds:

B + D = D => B = 0 not in agreement with (1) or (2) above so not possible
or B + D = 10 + D => B = 10 not possible
or B + D + 1 = D => B = -1 not possible
or B + D + 1 = D + 10 => B = 9

We have

A9A9A9
+ CDCDCD
--------------
1DDDDDF

1. If D = 0

A9A9A9
+ C0C0C0
--------------
100000F

But B ?F, so D can't be 0.

2. D?1 because E = 1

3. If D = 2

A9A9A9
+ C2C2C2
--------------
122222F

F can't be 1, because E ?F so D ?2

4. If D = 3

A9A9A9
+ C3C3C3
--------------
133333F
=> F = 2
=> A + C + 1 = 13
=> A + C = 12

Trying all possible combinations for A + C = 12:

A C
9 3 No
8 4
7 5
6 6 No
5 7
4 8
3 9 No

5. If D = 4

A9A9A9
+ C4C4C4
--------------
144444F
=> F = 3
=> A + C + 1 = 14
=> A + C = 13

Trying all possible combinations for A and C:

A C
9 4 No
8 5
7 6
6 7
5 8
4 9 No

6. If D = 5

A9A9A9
+ C5C5C5
--------------
155555F

=> A + C =14

A C
9 5 No
8 6
7 7 No
6 8
5 9 no

7. If D = 6

A9A9A9
+ C6C6C6
--------------
166666F

=> A + C = 15

A C
9 6 No
8 7
7 8
6 9 No

8. If D = 7

A9A9A9
+ C7C7C7
--------------
177777F

=> A + C = 16

A C
9 7 No
8 8 No
7 8 No

There are 12 solutions. These are all the solutions of the problem. They are written in the following table:

A B C D E F
8 9 4 3 1 2
7 9 5 3 1 2
5 9 7 3 1 2
4 9 8 3 1 2
8 9 5 4 1 4
7 9 6 4 1 3
6 9 7 4 1 3
5 9 8 4 1 3
8 9 6 5 1 4
6 9 8 5 1 4
8 9 7 6 1 5
7 9 8 6 1 5