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Too many threes and sixes (Posted on 2005-05-25) Difficulty: 2 of 5
The integer A consists only of 666 threes, and the integer B has only 666 sixes.

What digits appear in the product A x B ?

JUST PENCIL AND PAPER !!!

See The Solution Submitted by pcbouhid    
Rating: 4.0000 (5 votes)

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Solution Solution with proof | Comment 3 of 10 |
Instead of 666, let's write K. A number formed with K ones is (10^K-1)/9, so A=3.(10^K-1)/9 and B=6.(10^K-1)/9, and then AB= (2/9).(10^K-1).

(10^K-1)= 10^2K-2.10^K+1= 999...98000...01, a number with (K-1) nines, an eight, (K-1) zeroes, and a one.

Dividing by 9 produces 111...10888...89, which multiplied by 2 gives 222...21777...78: (K-1) twos, a one, (K-1) sevens, and an eight.


  Posted by Old Original Oskar! on 2005-05-25 18:29:13
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