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CTK Exchange
Ramsey2879
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Jun-11-05, 08:25 AM (EST) |
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1. "RE: 2 x ab x cde = fghi"
In response to message #0
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I will give some insight. It makes sense that either a or c = 1 in order that the product not be greater than 1000 since 2*40*300>9999 (assume that a,c <> 0 since a leading 0 is spurious). There are more choices if c = 1 but lets eliminate or check a=1 first. b and e can not equal 0 since then i= 0 = b would be a contradiction. We then check 2*13*4?? but find that this number is greater than 9876. Thus c must equal 3 if a = 1. Can you take it further from here? |
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sfwc
Member since Jun-19-03
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Jun-12-05, 11:23 AM (EST) |
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2. "RE: 2 x ab x cde = fghi"
In response to message #1
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>I will give some insight. It makes sense that either a or c
>= 1 in order that the product not be greater than 1000 since
>2*40*300>9999 (assume that a,c <> 0 since a leading 0 is
>spurious). There are more choices if c = 1 but lets
>eliminate or check a=1 first. b and e can not equal 0 since
>then i= 0 = b would be a contradiction. We then check
>2*13*4?? but find that this number is greater than 9876.
>Thus c must equal 3 if a = 1. Can you take it further from
>here?
Since 2 * 14 * 356 > 9876 we now have d = 0. Then b and e are both in the set {4, 6, 7, 8, 9}. Considering the problem modulo 9, we have 2(b + 1)(d + 3) ~ 3 - b - d, which rearranges to (2b + 3)(2d + 7) ~ 6. This has the 4 solutions (4, 7), (6, 9), (8, 4) and (9, 8), of which only (4, 7) yields a solution to the original problem:2 * 14 * 307 = 8596 The other case (c = 1) will have 8 cases to check; 4 for a = 3 and 4 for a = 4. None of these yields a solution, so that given is unique. Thankyou sfwc
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ramsey2879
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Jun-12-05, 07:20 PM (EST) |
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3. "RE: 2 x ab x cde = fghi"
In response to message #2
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I didn't finish this problem since I wanted to give the poster a chance to work on it. Help was asked for, not a solution. However, I liked your method of reducing the problem to a congruence equation. Note that since we already know that b = 4, the congruence equation should have been 2*5*(e+3) ~(sum i {i= 5 to 9})-e -> 2e ~ 5 + 9 -> e = 7 |
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