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# No. 1

### #11

Posted 10 November 2007 - 02:15 AM

### #12

Posted 15 November 2007 - 04:01 AM

### #13

Posted 30 December 2007 - 02:10 PM

### #14

Posted 12 February 2008 - 01:37 AM

(nothing in the instructions said that a number couldn't be the solution to more than one letter) (Duh)

### #15

Posted 27 September 2008 - 04:06 PM

I don't mean to necropost, but I only understood this up until (B-1) - E = A. How did you get A + F = B and F = 1 + E?Here's how I attempted it:

C=0 is quite obvious from the ***B - ***C = ***B.

From the same summation, it is seen that as C=0, F must be 5 as C carries from B and becomes 10 and then subtract that from 5 gets 5. so F=5.

As 1 was carried from B, so (B-1) - E = A and also A + F = B. So we know from this that F= 1 + E. Hence E = 4.

### #16

Posted 15 October 2008 - 09:29 PM

I don't mean to necropost, but I only understood this up until (B-1) - E = A. How did you get A + F = B and F = 1 + E?

You can deduce that A + F = B from the center column: DEFC + AB = DEBB. Since the hundred's place does not increase when you add A with F, A + F < 10, so A + F must equal B => (A + 5 = B). Using this result and substituting this value for B into: (B - 1) - E = A => ((A + 5) - 1) - E = A, eliminate the A's and you get E = 4.

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