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wu :: forums    riddles    medium (Moderators: SMQ, ThudnBlunder, towr, Icarus, Grimbal, Eigenray, william wu)    444445-digit number and not power of 2! « Previous topic | Next topic » Pages: 1  Reply Notify of replies Send Topic Print    Author  Topic: 444445-digit number and not power of 2!  (Read 694 times) gkwal Newbie     Posts: 25 444445-digit number and not power of 2!   « on: Jul 16th, 2007, 9:27pm » Quote Modify Each of the five-digit numbers from 11111 to 99999 (both inclusive) is written on a separate card (Clearly, there are 88889 such cards). Then, the cards are arranged in an arbitrary manner to form a chain. Prove that the 444445-digit number obtained in this way (note 444445 = 88889*5) is not equal to a power of 2. IP Logged towr wu::riddles Moderator Uberpuzzler Some people are average, some are just mean.     Gender: Posts: 13730 Re: 444445-digit number and not power of 2!   « Reply #1 on: Jul 17th, 2007, 1:26am » Quote Modify I think it's divisible by 3 IP Logged Wikipedia, Google, Mathworld, Integer sequence DB gkwal Newbie     Posts: 25 Re: 444445-digit number and not power of 2!   « Reply #2 on: Jul 17th, 2007, 8:41pm » Quote Modify sum of the numbers = (99999)*(99999+1)/2 - (11110)*(11111)/2     mod(sum, 3) =2     IP Logged Eigenray wu::riddles Moderator Uberpuzzler     Gender: Posts: 1948 Re: 444445-digit number and not power of 2!   « Reply #3 on: Jul 18th, 2007, 12:49am » Quote Modify Using a calculator, 1476411 < 444444 log2(10) < 444445 log2(10) < 1476415, so if the number is 2k, we can only have k = 1476412, 1476413, or 1476414.  However, mod 9, the number is congruent to 11111+...+99999 = 99999*100000/2 - 11110*11111/2, which is 8 mod 9.  But 2k = 8 mod 9 only when k = 3 mod 6, which none of the possibilities are. IP Logged towr wu::riddles Moderator Uberpuzzler Some people are average, some are just mean.     Gender: Posts: 13730 Re: 444445-digit number and not power of 2!   « Reply #4 on: Jul 18th, 2007, 1:32am » Quote Modify on Jul 17th, 2007, 8:41pm, gkwal wrote: sum of the numbers = (99999)*(99999+1)/2 - (11110)*(11111)/2     mod(sum, 3) =2     hmmm, yeah, maybe I should have actually calculated that. I suppose that would've been too simple.     Now if we only had numbers without 0's in them (which I tacitly overlooked), then perhaps...   Also interesting, if you take every number from 111111 to 999999 and concatenate them in whatever way, the result is divisible by 21. « Last Edit: Jul 18th, 2007, 1:41am by towr » IP Logged Wikipedia, Google, Mathworld, Integer sequence DB Eigenray wu::riddles Moderator Uberpuzzler     Gender: Posts: 1948 Re: 444445-digit number and not power of 2!   « Reply #5 on: Jul 18th, 2007, 1:56am » Quote Modify on Jul 18th, 2007, 1:32am, towr wrote: Also interesting, if you take every number from 111111 to 999999 and concatenate them in whatever way, the result is divisible by 21. In fact, if you concatenate the numbers from A=111...111 (n ones) to B = 999...999 (n nines), in any order, the result is divisible by A.   Edit: So it will never be a power of 2 (or 3, or 5, ...) « Last Edit: Jul 18th, 2007, 2:00am by Eigenray » IP Logged Pages: 1  Reply Notify of replies Send Topic Print Forum Jump: ----------------------------- riddles -----------------------------  - easy => medium   - hard   - what am i   - what happened   - microsoft   - cs   - putnam exam (pure math)   - suggestions, help, and FAQ   - general problem-solving / chatting / whatever ----------------------------- general -----------------------------  - guestbook   - truth   - complex analysis   - wanted   - psychology   - chinese « Previous topic | Next topic »

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