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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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 Paper Solution Comment 4 of 4 |

The paper solution is much more rewarding.  Not only do you see the pattern more readily coming to a general solution for n digit tridecimal multiplication of 4s and 6s, but you can also see the partial sums come together.

In this problem, specifically, you see how the carries from B (11) + C (12) = 23 [carry leave A (10)] turns each column's extra C into just an extra carry for the next column propegating until you arrive to the column with just C's and an one where the C's are turned into carries leaving just the one.  From there the carries reduce themselves until the last column has one carry giving the final 2.

This is a very elegant problem and only by working it out on paper do you see the underlying beauty.

The comment box seems to distort the equations, so I'll describe how to quickly create them.

Start with 4 times 6 by writting the 4 over the 6 leaving a line above for the carries.  For this problem you have only one sum of 1B and one carry of 1.

For the second one,, set up the problem in the same way with 44 over 66, but leave a space below the multiplicands for the carries from the sums.  This time the carry for the problem will be two 1s.  The partial sums will be 1CB and 1CB0 respectively, and they will have a single carry of 1.

For the third you have 444 over 666 and a carry of 111.  The partial sums are 1CCB, 1CCB0, and 1CCB00 with a carry of 1121.  Noticing the pattern yet?

The fourth one is 4444 over 6666 and a carry of four 1s.  The partial sums are 1CCCB, 1CCCB0, 1CCCB00, and 1CCCB000 with a carry of 112321.

Finally the fifth one is 44444 over 66666 and a carry of five 1s.  The partial sums are 1CCCCB, 1CCCCB0, 1CCCCB00, 1CCCCB000, and 1CCCCB0000 with a carry of 112344321.

By now, the pattern is pretty clear.

2 repeated (n-1) times & 1 & A repeated (n-1) times & B

So for 666 4s times 666 6s you would get a result with 665 2s followed by 1 followed by 665 As followed by B.

Solution:
22222222222222222222222222222222222222222222222222222222222222222222222222
22222222222222222222222222222222222222222222222222222222222222222222222222
22222222222222222222222222222222222222222222222222222222222222222222222222
22222222222222222222222222222222222222222222222222222222222222222222222222
22222222222222222222222222222222222222222222222222222222222222222222222222
22222222222222222222222222222222222222222222222222222222222222222222222222
22222222222222222222222222222222222222222222222222222222222222222222222222
22222222222222222222222222222222222222222222222222222222222222222222222222
22222222222222222222222222222222222222222222222222222222222222222222222221
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAB

Below are my equations, but I fear they will not line up properly.

`(1) (11)   111    1111      11111  4   44    444     4444      44444  6   66    666     6666      66666  -   --    ---     ----      -----(1)    (11)    (1121)  (112321)        (12344321) 1B  1CB   1CCB    1CCCB     1CCCCB 1CB0  1CCB0   1CCCB0    1CCCCB0 ---- 1CCB00  1CCCB00   1CCCCB00 21AB ------ 1CCCB000  1CCCCB000  221AAB -------- 1CCCCB0000   2221AAAB ----------     22221AAAAB`

 Posted by Joshua on 2010-08-10 09:55:42

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