Aha Problems in Egyptian Mathematics

Aha is an egyptian word meaning quantity. Aha problems were problems to compute an unknown quantity. Today these problems would be solved with algebra, but then they were solved using 'false position', a rhetorical method of solving certain algebra problems.

Several Papyri containing collections of arithemetic and aha problems, together with their solutions, have been found

Sample problem: A quantity and its fifth, added together, give 23. What is the quantity?

Solution by false position: Suppose the quantity is 5. Then add a fifth to get 6. Now do the same thing to 5 and 6 until 6 turns into 23. Like multiply 6 by . So the answer is 5= = .

Solution by algebra: Let x be the quantity. So x + x = 23. Solving for x gives x = 23.

Problems: Solve these problems from the Rhind papyrus, using false position. Check your work using algebra.

1. A quantity and its 1/2 added together become 16. What is the quantity?

2. A quantity and its 2/3 are added together. From the sum 1/3 of the sum is taken, leaving 10 remaining. What is the quantity?

3. Divide 100 loaves among 5 men in such a way that the shares received shall be in arithmetical progression, and that 1/7 of the sum of the largest three shares shall be equal to the sum of the smallest two. Find the shares.

Solutions

3. Let x be the smallest share and let y be the common difference. Then the sum of the shares is 5*x+10*y = 100. Also 1/7*(3*x+9*y) = 2*x + y

> eqns :={5*x+10*y = 100,1/7*(3*x+9*y) = 2*x + y};

> solve(eqns,{x,y});

By false position: Suppose the first two shares add to 10. Then 10 plus 7*10 adds to 80 instead of 100, so the sum of the first two shares is 100/80*10 = 100/8 = 12+1/2. Now twice the sum of the last 3 shares minus thrice the sum of the first 2 shares is

> 2*(87+1/2)-3*(12+1/2.);

But this is also 15 times the common difference. Suppose the common difference is 8. Then we get 120 instead, so the common difference is 137.5/120*8=137.5/15=9+1/6

> 137.5/15;

> x := (100/85); y := 8*x;

> sum(x+i*y,i=0..4); 1/7*(3*x+9*y)=2*x+y;

> x := 5/3; y := 55/6;

Double False Position

This method was developed to solve problems that 'single' false postition could not handle. Here is an example:

1. A purse of 100 dollars is to be divided amongst 4 men so that B may have 4 dollars more than A, and C 8 dollars more than B, and D twice as many as C. What is each man's share?

Solution by algebra: Let x be A's share. Then B, C, and D's share are x+4, x+12, and 2x+24 respectively. So x + x+4 + x+12 + 2x + 24 = 5x + 40 = 100. So x = 60/5 = 12.

x

Solution by false position: Suppose that x+e is guessed. Then the equation is x+e + x+e+4 + x+e+12 + 2(x+e) + 24 = 5(x+e) + 40 = 100 + 5e, so remove from our guess 1/5 the excess (if positive) or add to our guess 1/5 the deficit. Thus if you guess that A's share is 15, then the sum of the shares is 15 + 19 + 27 + 54 = 115. The correct answer is 15 - 15/5 = 12. Alternatively, if we remove the constant of 4+12+24 = 40 from both sides we get 75 on one side and 60 on the other, so the correct answer is 60/75*15 = 12

> 60/75*15;

> 15+ 19+27+54;

Solution by double false position: Suppose that A's share is guessed at x+e and also at x+f , giving a sum of shares at 100 + 5e and 100 + 5f respectively. Then compute ((x+e)5f - (x+f)5e)/(5f-5e) = (x5f - x5e)/(5f-5e) = x. Thus if we guess A's share is 15 and also guess that it is 20, giving share sums of 115 and 140 respectively, then the correct share for A is (15 (40)-20(15))/(40-15) = 20(15)/25 = 12.

> (15*(20)-20*(115))/(20-15) ;

Solve these problems by false position (single or double). Check your work by algebra.

1. Three men bought a boat for 10,000 dollars. The first man paid so much, the second that much plus 1000 more, and the last as much as the first two combined. How much did each man pay?

2. A man divided his estate amongst 3 sons and a daughter. To the oldest son he left 1/2 of his estate less 10,000 dollars. To the other two sons he left 1/4 and 1/5 of his estate respectively. His daughter received the rest, which was 500 dollars less than his least favored son. How much was his estate, and what was each share?