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 Post subject: Seven oaks question
PostPosted: Wed Aug 21, 2019 9:49 pm 
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Joined: Wed Jun 12, 2019 2:26 pm
Posts: 21
18)
In the sum below the letters J,M and C represent three different zero.
What is the value of J+M+C?

J J
M M
C C
---------
JMC


ANSWER IS 18


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 Post subject: Re: Seven oaks question
PostPosted: Wed Aug 21, 2019 10:26 pm 
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Joined: Wed May 09, 2007 2:09 pm
Posts: 1193
Location: Solihull, West Midlands
This is a fiddly one and I think it needs some trial and error

First notice that the three 2-digit numbers which are added together must all be in the 11 times table. So the sum must also be in the 11 times table

JJ = 11 x J
MM = 11 x M
CC = 11 x C

JMC = 11 x (J+M+C)

So the easiest way is to make a list of the first few three digit multiples of 11, add up the digits and see if they match the multiple

So 132 = 11 x 12 1+3+2 = 6 so this doesn't work
143 = 11 x 13 1+4+3 = 8

Try the next few until

198 = 11 x 18 1+9+8 = 18 RESULT!!


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 Post subject: Re: Seven oaks question
PostPosted: Fri Oct 18, 2019 10:20 am 
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Joined: Sun Aug 13, 2017 12:50 pm
Posts: 9
J J
M M
C C
---------
JMC

From the first column, it is clear that J+M=10
From the second column, it is clear that M=C+1
From the third column, it is clear that J=1

so , M=9, C=8
J+M+C=1+9+8=18


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