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 Post subject: Letters into digits
PostPosted: Wed Apr 28, 2010 9:09 pm 
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Joined: Tue Sep 01, 2009 3:38 pm
Posts: 2083
Location: Maidstone
In the following sums letter represent digits. No digit appears twice in any one sum. Find the value of each letter.

CD
CD
CD
CD+ = EC. Find the value of each letter C, D and E

LMN
LMN
LMN+ = NNN. Find the value of each letter L, M and N

PQ
PQ
QP+ = SSS. Find the value of each letter P, Q and S

Can you please explain the answer.

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 Post subject:
PostPosted: Wed Apr 28, 2010 9:33 pm 
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Only bit I can get so far is in the LMN one where N has to be the same digit at the end if multiplied by 3 - only one I can think of is 5

so answer NNN= 555

divide 555 by 3 = 185

so L=1 M=3 and N=5


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 Post subject:
PostPosted: Wed Apr 28, 2010 9:39 pm 
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In the first on C must be a multiple of 4 ie 4, 8, 12, 16, 20

so 2,4,6,8,0

If c is more than 2 then the whole sum will add up to more than 100 and the answer has 2 digits - so perhaps c=2, to get C=2 then d would have to be 3 and E would be 9?


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PostPosted: Wed Apr 28, 2010 9:44 pm 
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Location: Essex
1. C=2 D=3 E=9
2.L=1 M=8 N=5
3.0=4 P=8 Q=7 S=2

Are these the correct answers? I'm worse for wear and under the influence of a slosh of wine so my methods are pretty garbled (I worked backwards on the last one). Someone with a better idea than me will no doubt be along shortly!


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 Post subject:
PostPosted: Wed Apr 28, 2010 9:51 pm 
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Location: Maidstone
No I dont have answers its a scholarship sample paper.

How are you getting your answers? I am out of my breath :cry:

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 Post subject:
PostPosted: Wed Apr 28, 2010 10:55 pm 
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Is no 3:

p = 3
q = 4
s = 1

?


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PostPosted: Wed Apr 28, 2010 11:14 pm 
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Location: Maidstone
Minesatia, I dont know the answers, its just a sample scholarship paper I was send and unfortunately they dont give the answers :cry:

How did you get your answer. I am in complete darkness on how to work it out.

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PostPosted: Wed Apr 28, 2010 11:23 pm 
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Location: Essex
Told you I was a bit doolally! I added an extra variable to number 3 by copying it wrongly! No wonder it took so long! Minesatea looks right to me. :D


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PostPosted: Thu Apr 29, 2010 12:04 am 
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Location: Watford, Herts
Herman has explained the first two.

For the last one, the largest number you can get by adding three such numbers is 98+98+89 = 285, so S must be either 1 or 2, giving a sum of 111 or 222. The last column of the sum has 2 Qs and one P, so:

If S is 1, P must be odd and it can't be more than 5, i.e. it must be 1, 3 or 5, and we can consider the possible Qs that make the right column add up:
P=1, Q=5: 15+15+51 = 81 (no good)
P=3, Q=4: 34+34=43 = 111 (YES)
P=3, Q=9: 39+39+93 = 171 (no good)
P=5, Q=3: 53+53+35 = 141 (no good)
P=5, Q=8: 58+58+85 = 201 (no good)

If S is 2, P must be even and it can't be at least 6, i.e. it must be 6 or 8, and we can consider the possible Qs that make the right column add up:
P=6, Q=3: 63+63+36 = 162 (no good)
P=6, Q=8: 68+68+86 = 222 (YES)
P=8, Q=2: 82+82+28 = 192 (no good)
P=8, Q=7: 87+87+78 = 252 (no good)

So there are two solutions: (P=3, Q=4, S=1) and (P=6, Q=8, S=2).

Another approach is to rewrite the numbers being added as
10*P+Q + 10*P+Q + 10*Q+P = 21*P + 12*Q
and then consider all the possible Ps:

111 - 21*1 = 90 (doesn't divide by 12)
111 - 21*3 = 48 = 12*4
111 - 21*5 = 6 (doesn't divide by 12)

222 - 21*6 = 96 = 12*8
222 - 21*8 = 54 (doesn't divide by 12)


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 Post subject:
PostPosted: Thu Apr 29, 2010 9:09 am 
I am not sure a child would have sufficient time in an exam to trial many solutions.

I think for scholarship they are looking for a child with a very good sense of numbers.

In the third one, they would realise that adding three two digit numbers would give you answer below 300 so it must be 111 or 222.

If they assumed 111, they would then realise you couldn't have P being over 4. P can't be 4 or 2 because Q+Q+P must be an odd number ending in 1 which means Q+Q would have to be odd which is not possible. This leaves it having to be 3 which means Q has to be 4 (11-3 divided by 2) because 9 is fine to produce the units but is two large in the 10's.

On the first one, CD+CD+CD+CD =EC, someone with a good sense of numbers would know

a) C must be 1 or 2 but it can't be 1 because you can't add 4 of the same number and end up with an odd so it must be 2.
b) 4 times D must then be 12 (can't be 32 as that will result in a hundreds number) so D must be 3.

Basically you have to guide them towards a solution which would take no more than 5 minutes and it should be based on understanding of how numbers work rather than trial and error. Until recently I would have argued that you can't train up a scholarship winner -- they either have it in them or not -- but this year I had one who never managed any of the typical scholarship questions first time round but could faithfully do them after just one explanation.


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