Math Puzzle No. 1: Doubtful Doubling

Here's a puzzle that was first passed on to me by Al Lippert, who spent years providing the "Problem of the Day" at MathPath.  What goes after the last equals sign?


Doubtful Doubling: if 1+4 = 10, 2+8=20, 4+16=40, then 8+32=?




Hints: The answer we're looking for isn't 80.  And the equalities are all true!

Update: 11/6/20: read below for the solution.


There are undoubtedly many answers to the puzzle, but the solution we both came up with is: 130.

The hint with the puzzle said that all the equalities are true, and the answer was not 80.  How can this be?  The key is to write numbers in a different base!

We usually write numbers in base 10; starting from the right we have a 1's digit, then a 10's digit, then a 100's digit, and so on:

2345 = 2000 + 300 + 40 + 5 = 2 (1000) + 3 (100) + 4 (10) + 5(1).

In the puzzle above, the numbers on the left are written in base 10, but the numbers on the right are written in base 5, with a 1's digit, a 5's digit, a 25's digit, and so on.  Thus:

10 base 5 = 1(5) + 0(1) = 5 base 10

20 base 5 = 2(5) + 0(1) = 10 base 10

40 base 5 = 4(5) + 0(1) = 20 base 10

To answer the puzzle, we write 40 in base 5.  Because 40 = 25 + 15, the answer is 130:

130 base 5 = 1(25) + 3(5) + 0(1) = 40 base 10


Comments

azharhashmi said…
It is Such a good article
Keep Motivating us on giving these nice articles... Wolfram Mathematica

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