consider a six digit number that increases 6 times when it's last three digits are carried to the beginning of the number without their order being changed. then the largest digit in the given number is
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If we have a k-digit number with the first k-1 digits n and the last digit x, then we can write:
20n+2x=10k−1x+n20n+2x=10k−1x+n
Which simplifies to
19n=(10k−1−2)x19n=(10k−1−2)x
Therefore, because 1≤x≤91≤x≤9, 10k−1−210k−1−2 must be a multiple of 19 as x cannot provide that prime factor. Using modular arithmetic we can show that k must be a multiple of 18.
Additionally x cannot be 1 because that would result in n not being a k-1 digit number. The remaining choices for x will all work, however (it's easy to show this). Therefore there are infinite possible numbers which are all of the form
a1018b−119a1018b−119
For integers a, b satisfiying 2≤a<10,0
20n+2x=10k−1x+n20n+2x=10k−1x+n
Which simplifies to
19n=(10k−1−2)x19n=(10k−1−2)x
Therefore, because 1≤x≤91≤x≤9, 10k−1−210k−1−2 must be a multiple of 19 as x cannot provide that prime factor. Using modular arithmetic we can show that k must be a multiple of 18.
Additionally x cannot be 1 because that would result in n not being a k-1 digit number. The remaining choices for x will all work, however (it's easy to show this). Therefore there are infinite possible numbers which are all of the form
a1018b−119a1018b−119
For integers a, b satisfiying 2≤a<10,0
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