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# No. 1

15 replies to this topic

### #11 matfnch

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Posted 10 November 2007 - 02:15 AM

if u r Replaceing the same characters by the same numerals then how r u getting dat? if u do den u would get a=1 b=2 c=3 d=4... and so on rite?
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### #12 Cheryl C

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Posted 15 November 2007 - 04:01 AM

Whatever happened to simple factoring of polynomials? I must be in over my head. Can I get homework help here?
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### #13 whocheersbees

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Posted 30 December 2007 - 02:10 PM

This is cool. I am going to make my math teacher do this, it took me 2 hours to figure it out, since 6 AM when I decided to finally*Wink, wink* check the site out... I had to plug every # in every hole. I found out 9 wasn't used at all, LOL
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### #14 stupidamerican

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Posted 12 February 2008 - 01:37 AM

Maybe I missed something, but couldn't you just set every Letter to zero and say "DONE!!!"
(nothing in the instructions said that a number couldn't be the solution to more than one letter) (Duh)
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### #15 Izzy

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Posted 27 September 2008 - 04:06 PM

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.

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?
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### #16 pieater

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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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