Adding and Multiplying to Get 7.11Date: 01/13/99 at 20:10:20 From: Sam Subject: A+B+C+D=7.11 A*B*C*D=7.11 Here is the problem: Four numbers ($2.58 or $.14 - this kind of number) when added or multiplied equal $7.11. The numbers you use and the answer may only be in money units, with no rounding at all. I think you must use two values to get a divisor of 7.11. Thanks for any help you can give me, Sam Date: 01/14/99 at 11:35:42 From: Doctor Rob Subject: Re: A+B+C+D=7.11 A*B*C*D=7.11 Thanks for writing to Ask Dr. Math! This is fun, but a bit complicated. Express all the amounts of money in cents, instead of dollars. Call the amounts of money A, B, C, and D, which are all natural numbers. Then A + B + C + D = 711 A*B*C*D = 711000000 1 <= A < 711 1 <= B < 711 1 <= C < 711 1 <= D < 711 First observe that the arithmetic mean of A, B, C, and D is (A+B+C+D)/4 = 711/4 = 177.75 and their geometric mean is (A*B*C*D)^(1/4) = 711000000^(1/4) = 163.29 It is a useful fact that the geometric mean (g.m.) of any set of positive real numbers is always no greater than the arithmetic mean (a.m.). Now factor 711000000 into prime power factors A*B*C*D = 2^6 * 3^2 * 5^6 * 79 One of them must be divisible by 79. Say it is A that is divisible by 79. Then A = 79*a, where 1 <= a < 9, and a | 2^6 * 3^2 * 5^6. The fact that the arithmetic mean of B, C, and D is greater than their geometric mean gives (B+C+D)/3 = (711-79*a)/3 > (B*C*D)^(1/3) = [711000000/(79*a)]^(1/3) 493039*a^4 - 13312053*a^3 + 119808477*a^2 - 359425431*a + 243000000 < 0 This inequality has solutions 0.94197... < a < 4.04065... Because a is an integer, we have 1 <= a <= 4. Then B + C + D = 711 - 79*a <= 632 < 3125 = 5^5 so 5^5 cannot divide B, C, or D. Suppose first that B + C + D is not divisible by 5. Then exactly one of them is not divisible by 5, and we label this one B. If 625 = 5^4 divides either C or D, say C, then 625 <= C <= 711 < 2*625, so C = 625. Then B + D = 86 - 79*a >= 2, which implies a = 1, B + D = 7, and B*D = 14400. Then the a.m. of B and D is 3.5 and their g.m. is 120, which is impossible. Thus neither C nor D can be divisible by 5^4, so both C and D must be divisible by 5^3 = 125. Write C = 125*c, D = 125*d. Then we have B + 125*(c+d) + 79*a = 711, a*B*c*d = 9. Then c + d < (711-79-1)/125, so 2 <= c + d <= 5. Also a, B, c, and d can only be 1, 3, or 9, so one of c and d must be 1, say c, and then d is 1 or 3. Now a*B*d = 9, B + 125*d + 79*a = 632. If d = 1, then B + 79*a = 461 and a*B = 9, which lead to the quadratic equation 79*a^2 - 461*a + 9 = 0, which has no rational roots, a contradiction. If d = 3, then B + 79*a = 211 and a*B = 3, which lead to the quadratic equation 79*a^2 - 211*a + 3 = 0, which also has no rational roots, a contradiction. So B + C + D not divisible by 5 is impossible. Thus B + C + D must be divisible by 5, which implies that a = 4, A = 4*79 = 316, B + C + D = 395, B*C*D = 2250000. Now by an argument similar to the above, all three of B, C, and D must be divisible by 5, so put B = 5*b, C = 5*c and D = 5*d, which gives b + c + d = 79 b*c*d = 18000 = 2^4 * 3^2 * 5^3 As above, just two of b, c, and d are divisible by 5, and one of them by 5^2. Say b = 25*x, c = 5*y, 25*x + 5*y + d = 79, x*y*d = 2^4 * 3^2 = 144 Now x <= (79-5-1)/25, so 1 <= x <= 2. If x = 2, we get 5*y + d = 29, y*d = 72 and the a.m. of 5*y and d is 14.50, and their g.m. is 18.97, which is impossible. Thus x = 1, and 5*y + d = 54, y*d = 144 This leads to the quadratic equation 5*y^2 - 54*y + 144 = 0 (5*y - 24)*(y - 6) = 0 y = 6 or y = 24/5 The last root is not an integer, so it is extraneous, and you get the unique solution y = 6 x = 1 d = 24 c = 30 b = 25 a = 4 A = 316 B = 125 C = 150 D = 120 and the amounts of money were $3.16, $1.50, $1.25, and $1.20. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ Date: 05/17/2017 at 05:07:43 From: Ines Mueller Subject: Adding and Multiplying to Get 7.11 Dear Dr. Math, I have a very similar problem: find four amounts that when added or multiplied equal 7.35. I tried to solve my problem by referring to the 1999 conversation above, where the target was 7.11. I started with this: A + B + C + D = 735 (A + B + C + D)/4 = 735/4 = 183.75 A*B*C*D = 735000000 (A*B*C*D)^(1/4) = 164.65 But then factorization problems began ... I do not understand how Doctor Rob come to the seemingly important conclusion that "one of them must be divisible by 79." Nor do I understand the next formula. Would you assist me? Thanks in advance and greetings from Switzerland, Ines Date: 05/17/2017 at 23:21:36 From: Doctor Peterson Subject: Re: Adding and Multiplying to Get 7.11 Hi, Ines. The 79 comes from when Doctor Rob factored 711000000. He found that the largest prime factor is 79. Since that is a factor of A*B*C*D, it must be a factor of one of A, B, C, and D. In your case, you would want to factor your problem's analogous number, (735000000), and hope that it will provide a similarly notable prime (it doesn't). When you ask about "the next formula," do you mean this? The fact that the arithmetic mean of B, C, and D is greater than their geometric mean gives (B + C + D)/3 = (711 - 79*a)/3 > (B*C*D)^(1/3) = [711000000/(79*a)]^(1/3) = 493039*a^4 - 13312053*a^3 + 119808477*a^2 - 359425431*a + 243000000 < 0 Here, he is using the inequality "arithmetic mean > geometric mean" to write this: (B + C + D)/3 > (B*C*D)^(1/3) (711 - 79*a)/3 > [711000000/(79*a)]^(1/3) If we try solving this inequality, we might do this: (711-79*a)/3 > [711000000/(79*a)]^(1/3) (711-79*a) > 3*[711000000/(79*a)]^(1/3) (711-79*a)^3 > 27*711000000/(79*a) a(711-79*a)^3 > 27*711000000/79 243000000 - a(711-79*a)^3 < 0 This expands to 493039a^4 - 13312053a^3 + 119808477a^2 - 359425431a + 243000000 < 0 This is Doctor Rob's inequality. When I ask Wolfram|Alpha to solve this, it says 0.941974 < a < 4.04065 And that is just what he said. (I suspect he, too, used software to help!) Does this help you understand the work, so you can try using a similar method for your problem? (I can't promise that everything will work out the same with different numbers.) - Doctor Peterson, The Math Forum at NCTM http://mathforum.org/dr.math/ |
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