Remember the Math Adventurer’s Rule: Figure it out for yourself! Whenever I give a problem in an Alexandria Jones story, I will try to post the answer soon afterward. But don’t peek! If I tell you the answer, you miss out on the fun of solving the puzzle. So if you haven’t worked these problems yet, go back to the original posts. Figure them out for yourself—and then check the answers just to prove that you got them right.

## Egyptian Math in Hieroglyphs

(a) 5 + 7 = 12.

(b) 357 + 428 = 785, and the answer is written as:

(c) 200 – 75 = 125, written:

(d) 1000 – 163 = 837, written:

## Egyptian Math Puzzles

These are the answers to the Egyptian multiplication puzzles:

(e) 37 x 22 = 814, written:

(f) This problem was missing a few numbers, easily filled in by doubling the first line. 15 x 11 = 165, written:

(g) This fragment was missing the first two lines, making it more difficult to decipher. You needed to work backward from the first line shown to find the beginning of the calculation. 7 x 13 = 91, written:

Function machines

These were at the bottom of the multiplication puzzles page.

The first machine (first column of clues) used the function “triple the number and then add one,” or: $f\left(x \right) = 3x + 1$.

The second machine was a little harder. The function was “square the number and then take away two,” or: $f\left(x \right) = {x}^{2} - 2$.

## Another Egyptian Math Puzzle

The last hieroglyphic puzzle was the most difficult. The Egyptian scribe started with the number eight on his first line. He doubled it to get 16, then 32, then 64, then 128. Then he marked three lines and added them to get this total: $8 + 32 + 128 = 168$.

But that was not the answer to his problem. What the scribe really wanted to know was, “How many 8’s did I use to make 168?”

This hieroglyphic calculation is the division problem: $168 \div 8 = ?$

To find out his answer, the scribe must add the numbers in the left-hand column of the lines he had marked earlier: $1 + 4 + 16 = 21$ eights make 168.

Bonus story problem

And the answer to Deborah’s Horse Ranch puzzle: She had 6 stallions, 65 mares, and 14 foals, for a total of 85 horses. That’s a lot of stalls to muck out!

## Egyptian Geometry and Other Challenges

Our final set of Egyptian math puzzles featured geometry and algebra questions from the Rhind and Moscow papyri.

(1) If $A = \pi {r}^{2}$ is equal to ${\left[\frac{8}{9} \left(2r \right)\right]}^{2}$, then pi must be: $\pi = {\left(\frac{16}{9} \right)}^{2} \approx 3.16$.

(2) You will make the triangle with the greatest area when the two equal sides meet in a right angle.

(3) The width is 3 and the length is 4.

(4) The sides of the triangle are 4 and 10.

(5) If you count all the houses, cats, mice, heads of wheat, and measures of grain, they add up to 19,607 items. But if the mice ate the grain, and the cats ate the mice, then there are really only the houses and the cats left, right? That makes 56 in all.

The scribe’s “magic trick”

One last puzzle was to explain Scribe Ahmose’s magic trick from the Rhind papyrus. The easiest way to show that it will always work is to do the calculations in algebra:

• Tell your friend to think of a secret number.
(We will call the friend’s number $x$.)
• Then have him add 2/3 more to his number.
(Now the friend has $x + \frac{2}{3} x = \frac{5}{3} x$.)
• Finally, tell him to take away 1/3 of this total, and say the answer.
(That brings the friend up to $\frac{5}{3} x - \left( \frac{1}{3} \times \frac{5}{3} x \right) = \frac{10}{9} x$.)
• Now you must subtract 1/10 of that number to find the secret.
(We calculate $\frac{10}{9} x - \left( \frac{1}{10} \times \frac{10}{9} x \right) = \frac{9}{9} x = x$.)

## To Be Continued…

Read all the posts from the September/October 1998 issue of my Mathematical Adventures of Alexandria Jones newsletter.