(Send corrections to cbruni@uwaterloo.ca) Egyptian Matheatics Proble Set (i) Use the Egyptian area of a circle A = (8d/9) 2 to copute the areas of the following circles with given diaeter. d = 2. d = 3 3. d = 9/2. 256/8 2. 64/9 3. 6 (ii) Based on the tax rate in the lecture, copute the aount of grain needed to be taxed on the following plots of land. A plot of 0 setats. 2. A plot of four khet by two khet. 3. A circular plot of diaeter 8 khet (Hint: refer back to question!). 5 tons of grain 2. 2 tons of grain 3. 384 tons of grain (use (8d/9) 2 3/2) (iii) (Prograing) Write a progra that deterines what day of the year it is given the Egyptian calendar of alternating onths of 29 and 30 days. Can you factor in a leap year if a leap year is to happen only in years divisible by 5? Solution in C: # include <stdio.h> int ain ( void ) { # include <stdio.h> void day_of_year ( int n, int year ){ if (n <= 0) return printf (" -\n"); int cur = ; int axday = 0; for ( int i =0; i <2; i ++){ if (i%2 == ( year % 5 == 0 && i ==2)){
} } axday =30; } else { axday =29;} if (n <= axday ){ printf (" %.2 d /%.2 d /%.4 d\n",n,i+, year ); return ; } else { n -= axday ; } } printf (" -\n"); return ; return 0; (iv) Using Fibonacci s Greedy Algorith fro the notes, write each of the following fractions as a su of unit fractions. 4/7 2. 5/7 3. 5/7 4. 9/. 4/7 = /2 + /4 2. 5/7 = /2 + /5 + /70 3. 5/7 = /4 + /23 + /564 4. 9/ = /2 + /4 + /5 + /660 (v) The Egyptians tried really hard to ake the pyraids in the inverse ratio of the golden ratio. How well did they do? The actual easureents of the pyraids are 48.2ft high and 377.89ft for half the base length (the value we called b). Copute the value of the slant length a and copare b/a to the reciprocal of the golden ratio. a = h 2 + b 2 = (48.2) 2 + (377.89) 2 = 6.85 This gives b = 377.89/6.85 = 0.6762... a rearkably close to the reciprocal to the golden ratio! 2
(vi) Calculate the volue of the Giza pyraid fro the previous proble pyraid with diensions. If a future Pharoah wanted to create a pyraid that was 50% larger in size but still had b/a in ratio with the reciprocal of the golden ratio, what would the half base and height of such a pyraid be? V = 3 ((377.89)2 (48.2)) = 22905256.67ft 3 To ake it 50% larger, ultiply by.5 to get V new = 34357885.02 Solving in V new = 3 b2 h, h 2 + b 2 = a 2 and b/a = ϕ gives leading to (3V new /b 2 ) 2 + b 2 = ϕ 2 b 2 b = 6 9V 2 new (So the full base is 2 432.73 = 865.46) and ϕ 2 = 432.73 a = 700.7 h = a 2 b 2 = 550.442. (vii) Calculate the volue of a frustu with height 4, lower square base length of 5 and upper square base length of 3. How would such a forula look for a cone instead of a square based pyraid? V = 4 3 (32 + (3)(5) + 5 2 ) = 4 96 (9 + 5 + 25) = 3 3 (viii) Add 432 and 99 like an Egyptian! What difficulties to you notice fro the Egyptian syste? This converted becoes or in other words, 63. Notice that we need to write out each sybol the total nuber of ties and that 3
ll 9999 lll 9 lll lll lll 999 lll 99 lll ll l 9 99999 (ix) Write the fraction /7 as the su of 5 distinct unit fractions. (Hint: What happens when you add and )? n(n+) n+ We start by using Fibonacci s algorith to see that /2 is the largest unit fraction less than /7. Subtracting gives 7 2 = 5 34 Now, the largest unit fraction less than 5/34 is /7 and subtracting again yields 5 34 7 = 238 and so 7 = 2 + 7 + 238. However, this isn t the su of 5 unit fractions yet. We ake use of the fact that n(n + ) = n n + to write 6 = 2 3 and so 7 = 3 + 6 + 7 + 238. Repeating this with 2 = 3 4 gives 7 = 4 + 6 + 2 + 7 + 238 which copletes the requireents of the proble (note: there are any other solutions). 4
(x) Show that for all positive integers n > 2, the fraction 2 can be written as a su of two n non-zero unit fractions with different denoinators. Since = +, we see that n n(n+) n+ 2 n = 2 n(n + ) + 2 n +. The denoinators above are different (even when reduced to lowest ters) and this copletes the proof. (xi). Suppose a is a positive divisor of n 2. Let b = n 2 /a. Show that n = n + a + n + b We see that n + a + n + b = n + b + n + a (n + a)(n + b) 2n + a + b = n 2 + (a + b)n + ab = 2n + a + b 2n 2 + (a + b)n = 2n + a + b n(2n + (a + b)) since n 2 = ab = n copleting the proof. 2. Prove that for, n N coprie with 2 3n, the equation n = x + y with x, y N holds if and only if there exists a positive divisor a of n 2 such that ( ) (n + a), n + n2 and that x = n+a and y = n+n2 /a (or vice versa). a Solution - Nuber Theoretic: First, suppose that n = x + y. Let d = gcd(x, y), x = dx and y = dy for soe x, y N. Now, factor out /d fro the right and then rewrite the equation as dx y = n(x + y ). 5
Thus, y divides n(x + y ) and hence y n since gcd(y, x + y ) =. Siilarly, x n. Thus, let k N such that y k = n. Substituting into the above equation (with dx = x) gives x = k(x + y ) = kx + ky = kx + n. Now, fro y k = n, we see that k n and fro above, x n and so, kx n 2. Let a = kx. Then in the above, we have x = n + a and so n + a and further that x = n+a. Revisiting dx y = n(x + y ), we see that ultiplying both sides by k gives ya = dy x k = nk(x + y ) = na + ny k = na + n 2 and the left hand side is divisible by a and so y = n + n 2 /a giving that (n + n 2 /a) and y = n+n2 /a. In the converse direction, assue that a is a positive divisor of n 2 and that (n + a) and (n + n 2 /a). Then we have that (n + a)/ + (n + n 2 /a)/ = n + a + n + n 2 /a = (n + n2 /a + n + a) (n + a)(n + n 2 /a) = (a + 2n + n2 /a) n(n + n 2 /a + a + n) = n. Solution 2 - Analytic: In the forward direction, rewrite the given as n/ = x + y. Then, x > n/ and so there exists a natural nuber a such that x = n+a Siilarly, y = n+b for soe b N. Now, This gives n/ = n + a + n + b = (2n + a + b) (n + a)(n + b). 2n 2 + na + nb = n 2 + na + nb + ab and so n 2 = ab. Hence a n 2 and b = n 2 /a copleting the proof. The converse direction is siilar to the above. 3. Prove that for all n N, 3/n has a representation as a su of two different unit fractions if and only if there exists a prie factor of n not congruent to odulo 6.. 6
For the converse direction, assue that n is a product of pries all congruent to odulo 6. Then, by the previous exercise, we see that 3 n = x + y where 3x = n + a 0 (od 3) and 3y = n + n 2 /a 0 (od 3) (up to syetry). Now, by assuption, n (od 3) and any divisor of n ust contain only pries congruent to one odulo 6 and so, a (od 3). Thus, 0 n + a 2 (od 3), a contradiction. For the forward direction, we have three cases: Case : 3 n. In this case, the equation reduces to = + n x y works. and so, x = y = 2n Case 2: 2 n and 3 n. In this case, if n 2 (od 3), taking a = shows that n + a 0 (od 3) and n + n 2 /a 2 + 0 (od 3). However, if n (od 3), take a = 2 and then again the above equations hold. Thus, the previous exercise shows that 3/n ust have a representation as the su of two unit fractions. Case 3: 2 n, 3 n and p n for soe p 5 (od 6). If n (od 6), take a = p. If n 5 (od 6), then take a =. In either case, just like as above, we can use the previous exercise to see that 3/n can be expressed as the su of two unit fractions. 4. Show that 3/7 can t be expressed as a su of two unit fractions with different denoinators. Assue towards a contradiction that there exists distinct positive integers a and b such that 3 7 = a + b Rearranging gives 3ab = 7(a + b) and so WLOG, 7 a. Write a = 7k for soe positive integer k and so 3kb = 7k +b. Since b divides two of the ters, we see that b k. Write k = bl for soe positive integer l. Then 3bl = 7l +. As before, l and since all the ters were positive, we see that l =. Thus, 3k = 8 which is a contradiction. 7