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Topic: Delete A Digit, Get Power Of 2? (Read 356 times) |
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K Sengupta
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Delete A Digit, Get Power Of 2?
« on: Jul 27th, 2007, 8:07am » |
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Analytically determine all possible integers of the form 2p, where p is a positive integer such that if the first digit in the decimal expansion of 2p is obviated, the resulting number is equal to 2q, where q is a positive integer < p.
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Barukh
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Re: Delete A Digit, Get Power Of 2?
« Reply #1 on: Jul 27th, 2007, 11:03am » |
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32, 64 since p-q = 4?
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Eigenray
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Re: Delete A Digit, Get Power Of 2?
« Reply #2 on: Jul 31st, 2007, 4:24am » |
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We have 2p - d*10k = 2q, or
d*10k = 2q(2p-q-1)
Comparing odd parts gives x*5k = 2p-q-1, where x is either 1,3,5,7, or 9.
If x=1, 5, or 9, the LHS is 1 mod 4, while the RHS is 3 mod 4 unless p-q=1. But this would give x*5k=1, and k=0, which is invalid.
If x=3, then 2p-q=1 mod 3 implies p-q=2r must be even, and then 3*5k = 22r-1 = (2r+1)(2r-1). Since these two factors are odd and differ by 2, they are relatively prime, so they must be {1, 3*5k} or {3, 5k} in some order. Checking the 4 possibilities shows that the only way this can happen with k>0 is r=2, giving k=1. Since x=3 we have d=3 or d=6, and this gives the solutions 32 and 64.
If x=7, then 2p-q=1 mod 7 implies p-q=3r must be divisible by 3, and then
7*5k = 23r-1 = (y-1)(y2+y+1),
where y=2r. Since 1*(y2+y+1) - (y+2)*(y-1) = 3, and neither y-1 nor y2+y+1 can be divisible by 3, it follows these factors are relatively prime, so one of them is either 1 or 7, and checking the 4 possibilities shows no solutions with k>0.
So the only solutions are 32 and 64.
(Actually, this is a different proof than the one I used before.)
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