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 Power 2 The Cards (Posted on 2007-09-16)
88,889 cards are consecutively numbered from 11,111 to 99,999.

These cards are now arranged in a line in any arbitrary order.

Can the 444,445 digit number formed in this manner be a power of 2?

 See The Solution Submitted by K Sengupta Rating: 2.0000 (1 votes)

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 solution | Comment 1 of 2

Even at first glance it doesn't seem likely that powers of 2 could be formed:

There could only be 3 or 4 powers of two that have the requisite numbers of digits: call the first one k; the next ones are k*2, k*4, k*8 and k*16. The last one definitely has more digits than the first, and the next to last could have more digits than the first.

As a result, two factors weigh against the probability that one of them can be formed:

• The birthday problem: that among the 444,445 groups of 5 digits needed, there are no duplicates. Duplicates would require there be duplicates in the cards, and duplicates do not occur.
• Leading zeros: None of the cards contain leading zeros. It's not likely that of the 3 or 4 very long numbers there is one that has not even one zero in any of the 444,445 multiple-of-5-plus-1 positions.

Testing it:

The earliest 444,445-digit number is 10^444,444, with common logarithm 444,444. Divide by the common log of 2, to get the mixed-fractional power of 2 to get it:

`?444444/(log(2)/log(10)) 1476411.010204118871336876248519982527322069123641416705543103303 So the valid powers of 2 start at 1,476,412.`
`list   10   for I=1476412 to 1476415   20    L=(log(2)/log(10))*I-444444   30    print 10^L;"* 10^444444"   40   nextOKrun 1.98590399652092541238131509567199660313414381712026142157545981 * 10^444444 3.97180799304185082476263019134399320626828763424052284315091961 * 10^444444 7.94361598608370164952526038268798641253657526848104568630183919 * 10^444444 15.88723197216740329905052076537597282507315053696209137260367936 * 10^444444OK`

Indeed only 3 of these numbers have the required number of digits--the last one goes into one more digit.  (Note above, that the last 6 digits of each mantissa are probably not reliable, as we subtracted a 6-digit number from the common logarithm, but we won't be using those last few digits.)

Now divide the first few digits into groups of 5:

`  19859 03996 52092 54123 81315 09567 19966 03134 14381 71202  39718 07993 04185 08247 62630 19134 39932 06268 28763 42405  79436 15986 08370 16495 25260 38268 79864 12536 57526 84810  `

Already, within the first three cards of each, a leading zero is needed, in  03996, 07993 or 08370, and again, no actual card has a leading zero.

So it is indeed impossible that a power of 2 can be formed in this way.

 Posted by Charlie on 2007-09-16 16:55:08

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