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Topic: find middle digit (Read 2408 times) |
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inexorable
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find middle digit
« on: Jul 31st, 2011, 6:22pm » |
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A 7-digit number consists of 7 distinct digits (from 0-9). The product of the first 3 digits = product of central 3 digits = product of last 3 digits. Find the middle digit
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pex
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Re: find middle digit
« Reply #1 on: Aug 1st, 2011, 5:09am » |
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on Jul 31st, 2011, 6:22pm, inexorable wrote:| A 7-digit number consists of 7 distinct digits (from 0-9). The product of the first 3 digits = product of central 3 digits = product of last 3 digits. Find the middle digit |
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| hidden: | So we need to find a,b,c,d,e,f,g with abc=cde=efg, or equivalently, ab=de and cd=fg. We can quickly observe that only five nonzero numbers can be factored in two different ways with both factors under 10: ab=de and cd=fg are among
1*6 = 2*3 = 6
1*8 = 2*4 = 8
2*6 = 3*4 = 12
2*9 = 3*6 = 18
3*8 = 4*6 = 24
We need to pick two out of these five products, such that only one factor (d) appears in both. The only possibility is to let the products be 8 and 18, with d=2 being the required middle digit. |
One possible such number is 1892436, with 1*8*9 = 9*2*4 = 4*3*6 = 72.
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Grimbal
wu::riddles Moderator
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Re: find middle digit
« Reply #2 on: Aug 1st, 2011, 5:56am » |
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Same here
We need a,b,c,d,e,f,g such that abc = def = ghi = P.
We cannot have any 0 because P would be 0, but we cannot zero abc and ghi.
We cannot have 5 or 7 because that factor must appear in in all products but we have only 1 of each available.
Therefore we have the 7 digits 1,2,3,4,6,8,9.
abcdefg = 1*2*3*4*6*8*9 = 2^7*3^4
P^3 = abc*cde*efg = abcdefg*ce = 2^7*3^4*ce.
To make it a cube, we need two more factors 2 and 3. c*e = 2^2*3^2. (Larger powers are not possible since c*e<=8*9.
The only decomposition is c=4, e=9 (or the other way round).
Therefore c=4, e=9, P=2^3*3^2 = 72.
And this gives d = P/ce = 2. That is the middle digit.
You cold stop here, but just to check the solution indeed exists:
fg = 72/9 = 8, it can only be 1*8
ab = 72/c = 18, it is indeed 3*6, with the remaining digits.
The number 3642918 satisfies the conditions. And the middle digit is 2.
This being said, digit is the latin word for finger...
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danaz
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Re: find middle digit
« Reply #3 on: Aug 12th, 2011, 12:29am » |
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on Aug 1st, 2011, 5:09am, pex wrote:
| hidden: | So we need to find a,b,c,d,e,f,g with abc=cde=efg, or equivalently, ab=de and cd=fg. We can quickly observe that only five nonzero numbers can be factored in two different ways with both factors under 10: ab=de and cd=fg are among
1*6 = 2*3 = 6
1*8 = 2*4 = 8
2*6 = 3*4 = 12
2*9 = 3*6 = 18
3*8 = 4*6 = 24
We need to pick two out of these five products, such that only one factor (d) appears in both. The only possibility is to let the products be 8 and 18, with d=2 being the required middle digit. |
One possible such number is 1892436, with 1*8*9 = 9*2*4 = 4*3*6 = 72. |
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Beautiful explanation, could not have explained it any better  . I just did it through guess and check, got the same thing.
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