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# Problem solving watch

1. Hello

Any help on nailing this question using a quick approach pls? I tried to use the guessing method using different numbers, but that isn't how it is expected. So any hints or solutions would be great

2. was wrong sec
3. (Original post by Daniel Atieh)
Hello

Any help on nailing this question using a quick approach pls? I tried to use the guessing method using different numbers, but that isn't how it is expected. So any hints or solutions would be great

Alright, I'm here!
4. (Original post by raniafern)
Alright, I'm here!
perfect
5. (Original post by Daniel Atieh)
Hello

Any help on nailing this question using a quick approach pls? I tried to use the guessing method using different numbers, but that isn't how it is expected. So any hints or solutions would be great
I had a good think about this but I couldn't arrive to a proper method of solving this, but at first I posted that it's either 0 or 2, then I realised I'm wrong. The only mathematical thing I could think of using would be matrices but even that I'm unsure of how to use for this. 2 or 3 digits would be straight forward but 6 is something new for me. So I'll just sit tight for a proper solution until then.

For a pass code a 8 c d e f I got as far as:

6. (Original post by RDKGames)
I had a good think about this but I couldn't arrive to a proper method of solving this, but at first I posted that it's either 0 or 2, then I realised I'm wrong. The only mathematical thing I could think of using would be matrices but even that I'm unsure of how to use for this. 2 or 3 digits would be straight forward but 6 is something new for me. So I'll just sit tight for a proper solution until then.

For a pass code a 8 c d e f I got as far as:

Haha, yup. And then, you solve the equation by eliminating 10e + f, then you eliminate c and then put them together... You get 10A + 11D + C = 72, if I'm not mistaken... I've got to solve the rest. but Ethan Hunt is on TV at the moment
7. well by forming some equations and finding some restrictions you can at least shorten the trial and error procedure but I dunno if there's enough info to avoid that completely

For instance I am able to obtain
_ 8 _ _ 1 _
along with some equations
to deduce that either the first digit is 2 or it is 3, and checking out 2 that turns out to be correct giving

283517

I labelled it like so

a 8 b c d e

First thing to note is that 8 + c + e has to end in a 0. The only way this is possible is if c + e = 12.
Similarly b + e = 10
These lead to c - b = 2.

We have a8b + cde = 800, and since b + e = 10, we will carry the 1 and see that 8 + d + 1 = 10, giving d = 1.

We then note that a + c + 1 = 8, so a + c = 7.

Similarly we have a8 + bc + 1e = 80, and carrying the 2 from 8 + c + e = 20 we get a + b + 1 + 2 = 8, so a + b = 5

a cannot equal 1 because there is already a 1, and a cannot equal 4 because then b = 1, and there is already a 1. This leaves a =0,2,3 to check. a = 0 clearly doesn't work because then b = 5, and so e = 5 (as b + e = 10). a = 3 gives b = 2, which in turn gives c = 4, but then e = 8, which has already been used. So a = 2, from which everything else follows.

edit: oops, I forgot you can have 2,0 back at the start there. Still, the answer is right. (and 2,0 immediately falls apart, if e = 0 then b = 10, which can't be allowed, if e = 2 then b = 8 which has been used)

(Original post by Daniel Atieh)
Hello Any help on nailing this question using a quick approach pls? I tried to use the guessing method using different numbers, but that isn't how it is expected. So any hints or solutions would be great
8. (Original post by Daniel Atieh)
Hello

Any help on nailing this question using a quick approach pls? I tried to use the guessing method using different numbers, but that isn't how it is expected. So any hints or solutions would be great

You need to use some logic/deduction. Algebra is probably not the best approach.

A 8 + B C + D E = 80

A 8 B + C D E = 800

80 ends with '0' so you need 8 + C + E to end with a '0'. So C + E must end with a '2'.

So C, E could both be 1 but all the digits are different so that's not right.
Then C, E must add up to 12 since two single digit numbers can't add to 22.

So you could have (C,E) = (5,7), (7,5), (6,6), (8,4), (4,8), (9,3), (3,9).

Three of these are not possible because of duplicates.

Try carrying on from here, If you get stuck, post all your working / thoughts.

9. (Original post by RDKGames)
I had a good think about this but I couldn't arrive to a proper method of solving this, but at first I posted that it's either 0 or 2, then I realised I'm wrong. The only mathematical thing I could think of using would be matrices but even that I'm unsure of how to use for this. 2 or 3 digits would be straight forward but 6 is something new for me. So I'll just sit tight for a proper solution until then.

For a pass code a 8 c d e f I got as far as:

I'm not a fan of using these finicky 100s and stuff
best to just write it like the by hand addition we did in primary lol, intuitive and simpler
10. (Original post by 13 1 20 8 42)
well by forming some equations and finding some restrictions you can at least shorten the trial and error procedure but I dunno if there's enough info to avoid that completely
(Original post by notnek)
You need to use some logic/deduction. Algebra is probably not the best approach.
(Original post by raniafern)
Haha, yup. And then, you solve the equation by eliminating 10e + f, then you eliminate c and then put them together... You get 10A + 11D + C = 72, if I'm not mistaken... I've got to solve the rest. but Ethan Hunt is on TV at the moment
(Original post by RDKGames)
I had a good think about this but I couldn't arrive to a proper method of solving this, but at first I posted that it's either 0 or 2, then I realised I'm wrong. The only mathematical thing I could think of using would be matrices but even that I'm unsure of how to use for this. 2 or 3 digits would be straight forward but 6 is something new for me. So I'll just sit tight for a proper solution until then.
Thanks a bunch for taking part, and I saw interesting replies and approaches.
11. I solved it in less than a minute through quick trial and error.

You know that the ending digits of the latter two digit numbers have to equal 2 or 12 with means you only have to work with (0,2)(9,3)(7,5).

You can see instantly that you can't use 0 because it being the 4th digit would require the first to be 8 and it being the 6th would require the 3rd to also be 0.

You then know that digit 4 can't be 9 as that would make the second condition impossible and it takes seconds to test the reverse case and see that it wouldn't work as you would have to use 1 twice.

So now you simply pick one arrangement of 7 and 5 to give you the final answer.

I'm sure there is a far more elegant way of doing it, but this took no time at all really

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