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Power 2 The Cards (Posted on 20070916) 

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?

Submitted by K Sengupta

Rating: 2.0000 (1 votes)


Solution:

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No. The 444,445 digit number formed in this manner be can never be a power of 2.
EXPLANATION:
If possible, let the number correspond to the form 2^{m}, where m is an integer.
Then, it follows that:
10^{444,444} <= 2^{m} < 10^{444,445}
Or, 444,444*(log_{2}10) <= m < 444,445*log_{2}(10)
Or, 1476411.0102 <= m < 1476414.3321
Or, m = 1476412, 1476413 or 1476414; giving:
m (Mod 6) = 4, 5 or 0
Or, 2^m(Mod 9) = 7, 5 or 1. .........(*)
Now, 111111 + 111112 + ……+ 999999
= 999,999*500,000  (5555*11111), which is congruent to 8(Mod 9).
Therefore, it follows that the 444,445 digit number must be congruent to 8(Mod 9). This contradicts (*) since 2^m cannot correspond to 8(Mod 9).
Consequently, the 444,445 digit number formed in this manner be can never be a power of 2.

An alternative analysis and methodology has been posted by Charlie here and here.

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