DEFABC = 6(ABCDEF)
Date: 12/10/2001 at 06:39:38 From: Vincent Tan Subject: Real Numbers Let abcdef be a 6-digit integer such that defabc is 6 times the value of abcdef. Find the value of a + b + c + d + e + f.
Date: 12/10/2001 at 13:16:18 From: Doctor Greenie Subject: Re: Real Numbers Hello, Vincent - This is a very interesting problem - one that lends itself to two totally different methods of solution. Thanks for sending it to us. First approach - as a word arithmetic problem. One approach to this problem is as a word arithmetic problem, where the possible values for each digit are determined using logic and our knowledge of the process of multiplication. We have ABCDEF x 6 ------ DEFABC The first thing I notice here is that a 6-digit number multiplied by 6 yields a 6-digit number; this means A must be 1. So we have 1BCDEF x 6 ------ DEF1BC In the product, digit D must be 6, 7, 8, or 9. Looking at the multiplication process in the 100s column elminates two of these possibilities: if D is 7, then the "carry" from the 10s column to the 100s column would have to be 9, which is not possible; and if D is 9, then that same carry would have to be 7, which is also not possible. So D is either 6 or 8, and we have 1BC6EF 1BC8EF x 6 or x 6 ------ ------ 6EF1BC 8EF1BC In the first possibility, B must be 0, so we would have 10C6EF x 6 ------ 6EF10C The E in the product must be 2, 3, 4, or 5 (digits 0 and 1 are already used); looking at the multiplication process in the 1s and 10s column for each of these possible values for digit E shows that the only possible digits for E and " are E=3 and F=2; but that make C=2 also. The assumption that D=6 has led to a dead end, so we now know D is 8. So we have 1BC8EF x 6 ------ 8EF1BC Examining the "carry" from the 10000s column to the 100000s column, we see that B must be either 3 or 4. So we have 13C8EF 14C8EF x 6 or x 6 ------ ------ 8EF13C 8EF14C If B is 3, then, looking at the multiplication process in the 10000s column, we see that E must be either 0 or 2; but looking at the multiplication process in the 1s and 10s columns, we see that neither of these values is possible. So B is 4, and we have 14C8EF x 6 ------ 8EF14C Looking at the multiplication process in the 10000s column in this case, we see that E can be 5, 7, or 9. But trying each of these possible values for E in the multiplication process in the 1s and 10s column, we find that the only possible values for E and F are E=5 and F=7; these values make C=2. So we have our unique solution: 142857 x 6 ------ 857142 Second approach - as an algebra problem. The given problem says ABCDEF x 6 ------ DEFABC If we think of ABC and DEF as the 3-digits integers x and y, respectively, then ABCDEF is 1000x+y and DEFABC is 1000y+x; so we have 6(1000x+y) = 1000y+x 6000x+6y = 1000y+x 5999x = 994y 5999x 35x 5x y = ----- = 6x + --- = 6x + --- 994 994 142 Since x and y are integers, the fraction 5x/142 must be an integer; the smallest integer x for which this fraction is an integer is x=142, and we then have y = 6x + 5x/142 = 852+5 = 857 The solution to the algebraic problem is x=142 and y=857; so the solution to the original problem is (as before) 142857 x 6 ------ 857142 Thanks for a fun problem. Write back if you have any further questions about this. - Doctor Greenie, The Math Forum http://mathforum.org/dr.math/
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