ABCDEF is a 6-digit number where each of the digits is different.
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A, C, and E are odd and in increasing order.
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B, D and F are even.
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Any two adjacent digits differ by at least 2.
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The 2-digit numbers CD and EF are each a multiple of AB.
What is the 6-digit number?
From Mensa Puzzle Calendar 2022 by Fraser Simpson, Workman Publishing, New York. Puzzle for August 30.
The required 6-digit number is 183690
EXPLANATION
In each of AB, CD, and EF, we observe that the first digit must be odd and increasing and the second digit must be even.
If A=3, then for B>2, there is not more than one multiple of AB.
If AB=32, then 96 is the only valid multiple since for 64, the first digit is even. Contradiction.
Therefore A=1
If B=2, then the seemingly valid three multiples of AB are 36, 72, 96.
72 is invalid as the digit 2 has already been assigned to B.. Thus the only possible pairing is (CD, EF) = (36, 96), whereby we observe that D=F=6. Contradiction.
If B=4, then the valid multiples of AB are 56, 70, 98. Neither CD nor EF can correspond to 56, Or 98, since in this case we have consecutive letters being assigned consecutive digits. Contradiction. Therefore only 70 is valid and it is impossible to have any pairing of CD and EF,
If B=6, then the valid multiples of AB are 32, and 96. But 96 is not possible as the digit 6 has already been assigned to B. With the sole multiple 32 it is impossible to effect any pairing
If B =8, then the valid multiples of 18 are 36. 54, 72, 90.
Neither CD nor EF can be 54 which contains consecutive digits.
Then, the possible pairings are:
(36, 72), (36, 90), (72, 90)
But (CD,EF) = (36,72) would imply that D and E are assigned consecutive digits 6 and 7. This is a contradiction.
(CD, EF) = (72,90) would imply that B and C are assigned the consecutive digits 8 and 7. This is a contradiction.
Accordingly, (CD, EF) = (36, 90)
Consequently, 183690 is the required 6-digit number.
Edited on September 7, 2022, 7:20 am
Puzzle Solution
