Solving a simple addition / multiplication puzzle.

02172024, 09:44 PM
(This post was last modified: 02182024 01:43 AM by matalog.)
Post: #1




Solving a simple addition / multiplication puzzle.
My dad forwarded me an email I had sent him in 2002:
"A customer at a 711 store selected four items to buy, and was told that the cost was $7.11. He was curious that the cost was the same as the store name, so he inquired as to how the figure was derived. The clerk said that he had simply multiplied the prices of the four individual items. The customer protested that the four prices should have been ADDED, not MULTIPLIED. The clerk said that that was OK with him, but, the restult was still the same: exactly $7.11. What were the four prices?" So, obviously we have a+b+c+d =7.11= a*b*c*d. I must have took the time to work it out on paper back then, but now I have my HP Prime. To be honest, I usually use this calculator to do something different regarding connectivity or fast processing on a mobile computer, but I rarely use it for these type of real world problems and need some help. Is there a way to input those 2 equations, and get a solve app or command on the HP Prime to return the answers? I suppose that the answers will be 2dp each, and not more. I tried this in CAS Code: solve( { a+b+c+d=7.11, a+b+c+d=7.11}, {a, b, c, d}) 

02182024, 03:15 AM
(This post was last modified: 02202024 07:27 PM by Albert Chan.)
Post: #2




RE: Solving a simple addition / multiplication puzzle.
711 = 79 * 9 > a = 0.79k, for positive integer k
b+c+d = 7.11  0.79k k*b*c*d = 9. 9*10^6, divisors ≥ 100: 100,120,125,144,150,160,180,192,200,225,240,250,288,300, ... k=1: b+c+d = 6.32 b*c*d = 9 9 ≈ 3.00 * 1.732 * 1.732 > sum of factors ≈ 6.46, too high 9 ≈ 2.88 * 1.768 * 1.768 > sum of factors ≈ 6.42, again too high 9 ≈ 2.50 * 1.897 * 1.897 > sum of factors ≈ 6.29, possible 9 = 2.50 * 2.00 * 1.80 > sum of factors = 6.30 9 = 2.50 * 2.25 * 1.60 > sum of factors = 6.35 > 2.5 no good 9 ≈ 2.40 * 2.25 * 1.667 > sum of factors ≈ 6.317 9 ≈ 2.40 * 2.40 * 1.563 > sum of factors ≈ 6.363 > 2.4 no good 9 ≈ 2.25 * 2.25 * 1.778 > sum of factors < 6.317, can't get higher, k=1 no good We still do sum of factors, even for fractional pennies, so that we know when to stop. k=4 give the first solution: 3.16 + 1.50 + 1.25 + 1.20 = 7.11 3.16 * 1.50 * 1.25 * 1.20 = 7.11 

02182024, 03:49 AM
Post: #3




RE: Solving a simple addition / multiplication puzzle.
Albert you are a very clever fellow!
On the face of it, there are four variables but only three pieces of information, being three pairs of quantities that are equal! Can you explain a little more about the insight that you have that allows us to get past that? What is the significance of your first line? 

02182024, 04:36 AM
Post: #4




RE: Solving a simple addition / multiplication puzzle.
Before solving puzzle, I checked if AM ≥ GM
AM(a,b,c,d) = 7.11/4 ≈ 1.78 GM(a,b,c,d) = ^{4}√(7.11) ≈ 1.63 AM ≥ GM > solution is possible. (02182024 03:49 AM)Johnh Wrote: What is the significance of your first line? If we use units of pennies, we are really searching for integer solution. A + B + C + D = 711 A * B * C * D = 711 * 10^6 = 79 * 9 * 10^6 79 is a prime number, must go in one of the variable, I just put it in A B + C + D = 711  79k k * B * C * D = (9 * 10^6) (B, C, D) must be divisors of (9 * 10^6) 

02182024, 09:07 AM
(This post was last modified: 02182024 10:11 AM by Thomas Klemm.)
Post: #5




RE: Solving a simple addition / multiplication puzzle.
This Python program is based on 2.1.5 Our Third Additive Solution of The 711 Problem:
Code: N = 7_11_00_00_00 Result 120 125 150 316 Using a single listcomprehensions makes it maybe easier to read: Code: [ Result [(120, 125, 150, 316)] In this paper you can also find 2.2 Mathematical Approach: Quote:We could also attempt to solve this without a computer program, and You can search for a+b+c+d=7.11=a*b*c*d and find plenty solutions. 

02182024, 02:43 PM
(This post was last modified: 02232024 06:21 PM by Albert Chan.)
Post: #6




RE: Solving a simple addition / multiplication puzzle.
(02182024 03:15 AM)Albert Chan Wrote: 711 = 79 * 9 > a = 0.79k, for positive integer k Instead of guessing, we may first find out where to look. (I skipped k=1..3) k=4 > a = 0.79k = 3.16 b+c+d = 3.95 > AM ≈ 1.3167 b*c*d = 2.25 > GM ≈ 1.3104 AM > GM, and they are *very* close! Let's assume (b,c,d) evenly distributed, b=cx, d=c+x c ≈ AM ≈ 1.3167 b*c*d = c*(c^2x^2) = 2.25 > x ≈ 0.1577 > (b,c,d) ≈ [1.1590, 1.3167, 1.4744] > c must be within extremes, possibilities = 1.20, 1.25, 1.44 Lets try the more likely center first, c = 1.25 (b+d) = 3.95  1.25 = 2.7 (b*d) = 2.25 / 1.25 = 1.8 (bd)² = (b+d)²  4*(b*d) = 2.7²  4*1.8 = 0.09 = 0.3² (b, d) = ((b+d) ± (bd))/2 = (2.7 ± 0.3)/2 = (1.20, 1.50) > (a, b, c, d) = 3.16, 1.20, 1.25, 1.50 Another way is assume (b,c,d) skewed extremely to one side, i.e. c = d Cas> solve(b+d+d=3.95 and b*d*d=2.25, [b,d]) \(\left(\begin{array}{cc} 1.1390 & 1.4055 \\ 1.5028 & 1.2236 \\ 5.2583 & 0.6541 \end{array}\right)\) max(b,d) = (1.4055, 1.5028), possibilities = 1.44, 1.50 min(b,d) = (1.1390, 1.2236), possibilities = 1.20 Solving cubic is harder, but we reduced to test only for b = 1.20 With only 2 possible d's left, we could try them all to get c. 2.25 / (1.20*1.44) = 1.30208... ✘ 2.25 / (1.20*1.50) = 1.25 3.16 + 1.20 + 1.25 + 1.50 = 7.11 ✔ 

02182024, 03:34 PM
Post: #7




RE: Solving a simple addition / multiplication puzzle.
(02182024 04:36 AM)Albert Chan Wrote: Before solving puzzle, I checked if AM ≥ GM Could you explain why 79 must be one of the values? I cannot find the answer. Thank you Patrick 

02182024, 11:51 PM
Post: #8




RE: Solving a simple addition / multiplication puzzle.
(02182024 03:34 PM)case2001 Wrote: Could you explain why 79 must be one of the values? I cannot find the answer. A * B * C * D = 711 * 10^6 = 79 * 9 * 10^6 If we divide both side by prime 79, RHS still is integer, so does LHS. (A * B * C * D) / 79 = 9 * 10^6 This mean one of the variables must be divisible by 79, to make LHS match RHS (02172024 09:44 PM)matalog Wrote: Is there a way to input those 2 equations, and get a solve app or command on the HP Prime to return the answers? I don't know about HP Prime, but Python z3 solver can solve the puzzle in seconds. >>> from z3 import * >>> A, B, C, D = Int('A'), Int('B'), Int('C'), Int('D') >>> solve(A+B+C+D == 711, A*B*C*D == 711*10**6, A>0, B>0, C>0, D>0) [D = 120, C = 316, B = 125, A = 150] 

02192024, 01:24 AM
Post: #9




RE: Solving a simple addition / multiplication puzzle.
Thank you


02192024, 04:53 PM
(This post was last modified: 02242024 02:47 AM by Albert Chan.)
Post: #10




RE: Solving a simple addition / multiplication puzzle.
(02182024 04:36 AM)Albert Chan Wrote: Before solving puzzle, I checked if AM ≥ GM It is a good idea to maintain AM/GM inequality thorughtout, to reduce search space. Example, if we assume b=c=d, and have AM = GM: Cas> solve((7.11a)/3 = (7.11/a)^(1/3), a) → [0.744, 3.192] Cas> solve((7.113*d) = 7.11/d^3, d) → [1.306, 2.122] Or, we solve both together Cas> solve(a+d+d+d = 7.11 and a*d*d*d = 7.11, [a,d]) \(\left(\begin{array}{cc} 3.192 & 1.306 \\ 0.744 & 2.122 \end{array}\right)\) Add back AM/GM inequality, with (a,b,c,d) sorted, unit = pennies a = 75 .. 130 d = 213 .. 319 Code: N = 711 * 10**6 120 125 150 316 We could also loop a and d, and solve (b,c) with quadratic formula. But trial and errors with all possible b's probably just as quick. Comment: Do this by hand, I would probably try A in reverse order. Even if tested number is not a, it may have hit b or c. >>> A[::1] [125, 120, 100, 96, 90, 80, 79, 75] 

02202024, 06:47 PM
(This post was last modified: 02202024 08:05 PM by Albert Chan.)
Post: #11




RE: Solving a simple addition / multiplication puzzle.
(02192024 04:53 PM)Albert Chan Wrote: Do this by hand, I would probably try A in reverse order. Perhaps we should *not* check if the number hit b or c, to avoid double counting. Try first number from A[::1], apply AM/GM inequality for (b, d) valid ranges: a = 125 b = 094 .. 133, possibilities = [125, 120, 100] d = 247 .. 318, possibilities = [250, 288, 300, 316] It is true that from possible (b,d)'s, we have solved the puzzle. What if solution is not here? Then, we try the next one on the list. a = 120 b = 098 .. 136, possibilities = [125, 120, 100] d = 247 .. 318, possibilities = [250, 288, 300, 316] Now, (a,b)=(120,125) is double counted. Same thing happens for case a=100 It is better to assume a is really the smallest. For a=125, only test for b=125 With a≤b restriction, no double counting. Code: a b b's d d's (b's)×(d's) Use quadratic formula for (c,d), cases to check = 17 But even tried all possible (c,d)'s, total cases is only 48 What if we know a is multiple of 79? (Again, (b,c,d) assume sorted) Code: a b b's d d's (b's)×(d's) Note: we don't have to actually solve all the valid ranges. Keeping same GM, if numbers are further apart, AM will grow bigger. If it is impossible to make AM ≥ GM, there is no solution ... we skipped it. This allowed "solving" AM/GM inequality by trial and errors. (see post #2) 

02222024, 09:47 PM
(This post was last modified: 02222024 11:37 PM by deachp.)
Post: #12




RE: Solving a simple addition / multiplication puzzle.
Hello,
It's an underdetermined system of 4 variables and two equations, therefore it has infinitely many solutions. If the goal is to obtain positive values with two decimal points, one can combine a quadratic equation, divisor calculation, and filtering out negative numbers and rational numbers. Here, I've developed a program for the 50G calculator that could be ported to the HP Prime: « 711 0 0 0 0 0 0 > P a b c d ldivis ldivis2 « 4 P * 1000000 * DIVIS 'ldivis' STO 1 ldivis SIZE FOR i ldivis i GET 'a' STO 4 P * a / DIVIS 'ldivis2' STO 1 ldivis2 SIZE FOR j ldivis2 j GET 'b' STO IF '((a+b)P)^2' >NUM '4*P*1000000/(a*b)' >NUM >= THEN IF '((P(a+b))+SQRT(((a+b)P)^24*P*1000000/(a*b)))/2' >NUM DUP FP 0 == THEN 'c' STO 'Pcab' EVAL 'd' STO IF a 0 > b 0 > AND c 0 > AND d 0 > AND THEN a I>R 100 / "a" >TAG b I>R 100 / "b" >TAG c I>R 100 / "c" >TAG d I>R 100 / "d" >TAG END ELSE DROP END END NEXT NEXT » » SQRT: Square root symbol > : Arrow a:1.25 b:3.16 c:1.5 d:1.2 It takes 20 sec., to emulator in "Authentic Calculator Speed" mode, to found the answers. It can be optimized. Dante Aroní C. 

02242024, 03:20 AM
Post: #13




RE: Solving a simple addition / multiplication puzzle.
a*b*c*d = 79 * 3^2 * 10^6
This implied one and only one variable is divisible by 79 Also, at least one variable only divisible by prime 2 or 5 We loop through these 2 variables to get the others. From AM/GM inequality, variables within 75 .. 319 (see post #10) Code: A = [79,158,237,316] 316 125 120.0 150.0 

02242024, 05:49 PM
(This post was last modified: 02252024 04:45 PM by Albert Chan.)
Post: #14




RE: Solving a simple addition / multiplication puzzle.
CMS Vol 32 #5, Sep06, page 12: M203 Mayhem solution
The solution skipped trial and errors, and started with a = 79*4 = 316 Here is another solution, using simple parity check. To get sum of 711, even a, (b+c+d) is odd, we need an odd number. Odd variable possibilities, b = [75, 125, 225] Based on AM/GM inequality, variables within 114 .. 150 > b = 125 (see post #11) Assuming we don't know that, it is still simple to check them all a = 316: b = 75 > (c+d)=320, (c*d)=30000 > (cd)² = 17600 < 0 b = 125 > (c+d)=270, (c*d)=18000 > (cd)² = 900 = 30² > (c,d)=(270±30)/2 b = 225 > (c+d)=170, (c*d)=10000 > (cd)² = 11100 < 0 > (a, b, c, d) = (316, 125, 150, 120) Another way is to figure out how 5^6 factor get splitted among (b,c,d). 5^4 = 625 > 319, outside AM≥GM criteria, we are down to 2 choices. 5^6 = 5^(3+2+1) 5^6 = 5^(2+2+2) (b+c+d) = 711  79k = 79*(9k) If k=4, RHS divisible by 5, but still not divisible by 25 > a = 79*4 = 316, b = (125 or 250) are the only cases to check. Again, assume we don't know (b,c,d) valid range = 114 .. 150 > b = 125 If b = 250, AM(c,d) = (711ab)/2 = 72.5, GM(c,d) = √(711e6/a/b) = √9000 ≈ 94.9 AM < GM, impossible. If b = 125, AM(c,d) = (711ab)/2 = 135, GM(c,d) = √(711e6/a/b) = √18000 ≈ 134.2 AM > GM: (c,d) = AM ± √(AM^2  GM^2) = 135 ± √225 = 135 ± 15 > (a, b, c, d) = (316, 125, 150, 120) 

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