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 Too many fours and sixes (Posted on 2010-08-07)
Each of A and B is a tridecimal (base 13) positive integer, where A is formed by writing the digit 4 precisely 666 times and, B is formed by writing the digit 6 precisely 666 times.

Determine the distinct digits that appear in the tridecimal representation of the product A*B.

Note: For an extra challenge, solve this problem using only pencil and paper.

 No Solution Yet Submitted by K Sengupta Rating: 3.5000 (2 votes)

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 Solution | Comment 2 of 4 |

An n-digit base 13 repunit (number whose digits are all 1) has a value of (13^n-1)/12.  Then the base 13 number of 666 4s is 4*(13^666-1)/12 = (13^666-1)/3 and the number of 666 6s is 6*(13^666-1)/12 = (13^666-1)/2.

Their product is (13^666-1)^2/6.  This can be expressed as 13^666*[2*(13^666-1)/12] - [2*(13^666-1)/12]

The number in brackets is 666 2s in base 13, and 13^666 is a 1 followed by 666 0s.  Then the subtraction has a first number of 666 2s followed by 666 0s and the second number is 666 2s, when both are expressed in base 13.

If we let a=10 and b=11 for base 13 digits, then the difference is 665 2s, followed by one 1, followed by 665 a's, followed by one b.  The distinct digits are 1,2,a,b.

 Posted by Brian Smith on 2010-08-07 23:13:19

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