Combining NumbersDate: 08/04/2002 at 15:36:00 From: Roger Subject: A problem from my daughter's tutor! Hi, Combine the numbers 3, 3, 8, and 8, using only addition, subtraction, multiplication, and division, to come up with the number 24. We appear to have used every combination and have not come up with the answer 24. We just need to know how it is worked out please. Warm regards, Roger Date: 08/05/2002 at 18:14:19 From: Doctor Ian Subject: Re: A problem from my daughter's tutor! Hi Roger, What makes this kind of problem so difficult is that it's easy to just start trying things, without any good way of keeping track of what you've tried, or any way of knowing when you've tried everything. Ultimately, all the operators (+, -, *, /) are binary, which means they have to work on two operands, so we have to begin by choosing two operands to work with. There are three ways we can do that: (3,3,8,8) -> ((3,3),8,8) ((3,8),3,8) ((8,8),3,3) Here are the results we can get by applying one operator to these pairs of operands: (a,b) a+b a-b b-a a*b a/b b/a Remaining (3,3) -> 6 0 0 9 1 1 (8,8) (3,8) -> 11 -5 5 24 3/8 8/3 (3,8) (8,8) -> 16 0 0 64 1 1 (3,3) Note that we can use each of the numbers in the center as a new starting point for a smaller version of the same problem: (3,3,8,8) -> ((3,3),8,8) => (6,8,8) (0,8,8) (9,8,8) (1,8,8) ((3,8),3,8) => (11,3,8) (-5,3,8) (5,3,8) (24,3,8) (3/8,3,8) (8/3,3,8) ((8,8),3,3) => (16,3,3) (0,,3,3) (64,3,3) (1,3,3) Now we can do the same thing (i.e., pair the remaining operands in various ways) again: (3,3,8,8) -> ((3,3),8,8) => (6,8,8) -> ((6,8),8) ((8,8),6) (0,8,8) -> ((0,8),8) ((8,8),0) (9,8,8) -> ((9,8),8) ((8,8),9) (1,8,8) -> ((1,8),8) ((8,8),1) ((3,8),3,8) => (11,3,8) -> ((11,3),8) ((11,8),3) ((3,8),11) (-5,3,8) -> ((-5,3),8) ((-5,8),3) ((3,8),-5) (5,3,8) -> ((5,3),8) ((5,8),3) ((3,8),5) (24,3,8) -> ((24,3),8) ((24,8),3) ((3,8),24) (3/8,3,8) -> ((3/8,3),8) ((3/8,8),3) ((3,8),3/8) (8/3,3,8) -> ((8/3,3),8) ((8/3,8),3) ((3,8),8/3) ((8,8),3,3) => (16,3,3) -> ((16,3),3) ((3,3),16) (0,3,3) -> ((0,3),3) ((3,3),0) (64,3,3) -> ((64,3),3) ((3,3),64) (1,3,3) -> ((1,3),3) ((3,3),1) For the next step, we need to make up another table for the distinct pairs in the rightmost column: (a,b) a+b a-b b-a a*b a/b b/a Remaining (-5,3) -> -2 -8 8 -15 -5/3 -3/5 8 (-5,8) -> -3 -13 13 -40 -5/8 -8/5 3 (0,3) -> 3 -3 3 0 0 ~ 3 (0,8) -> 8 -8 8 0 0 ~ 8 (3/8,3) -> 27/8 -21/8 21/8 9/8 3/24 8 8 (3/8,8) -> 67/8 -61/8 61/8 3 3/64 64/3 3 (1,3) -> 4 -2 2 3 1/3 3 3 (1,8) -> 9 -7 7 8 1/8 8 8 (8/3,3) -> 17/3 -1/3 1/3 8 8/9 9/8 8 (8/3,8) -> 32/3 -16/3 16/3 64/3 8/24 3 3 (3,3) -> 6 0 0 9 1 1 0, 1, 16, 64 (3,8) -> 11 -5 5 24 3/8 8/3 -5, 3/8, 8/3, 5, 11, 24 (5,3) -> 8 2 -2 15 5/3 3/5 8 (5,8) -> 13 -3 3 40 5/8 8/5 3 (6,8) -> 14 -2 2 48 6/8 8/6 8 (8,8) -> 16 0 0 64 1 1 0, 1, 6, 9 (9,8) -> 17 1 -1 72 9/8 9/8 8 (11,3) -> 14 8 -8 33 11/3 3/11 8 (11,8) -> 19 3 -3 88 11/8 8/11 3 (16,3) -> 19 13 -13 48 16/3 3/16 3 (24,3) -> 27 21 -21 72 8 1/8 8 (24,8) -> 32 16 -16 192 3 1/3 3 (64,3) -> 67 61 -61 192 64/3 3/64 3 Okay, so what can we do with this? Well, the _only_ way we can end up with 24 is if we can do it in one by adding, subtracting, multiplying, or dividing something from the Remaining column in a given row with something else in the row. For example, let's look at the row (a,b) a+b a-b b-a a*b a/b b/a Remaining (-5,3) -> -2 -8 8 -15 -5/3 -3/5 8 There is no way to combine 8 with -2, -8, 8, -15, -5/3, or -3/5 to end up with 24. So we can forget about this row. Let's cherry-pick the easy ones: (a,b) a+b a-b b-a a*b a/b b/a Remaining (-5,3) -> -2 -8 8 -15 -5/3 -3/5 8 (No) (-5,8) -> -3 -13 13 -40 -5/8 -8/5 3 (No) (0,3) -> 3 -3 3 0 0 ~ 3 (No) (0,8) -> 8 -8 8 0 0 ~ 8 (No) (3/8,3) -> 27/8 -21/8 21/8 9/8 3/24 8 8 (No) (3/8,8) -> 67/8 -61/8 61/8 3 3/64 64/3 3 (No) (1,3) -> 4 -2 2 3 1/3 3 3 (No) (1,8) -> 9 -7 7 8 1/8 8 8 (No) (8/3,3) -> 17/3 -1/3 1/3 8 8/9 9/8 8 Yes! We have a winner: 8 divided by 1/3 is 24. Now, how did we get here? Tracing back, we find that the corresponding expression is 8 8 ------- = ----- = 8*3 = 24 3 - 8/3 1/3 So, this wasn't pretty, but I knew that I was covering all the bases, and that I wasn't touching any of them more than once. It's nice when you can 'see' the answer directly, which sometimes happens; but when that doesn't work, it's good to have a backup plan. - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ Date: 08/06/2002 at 16:20:56 From: Roger Percival Subject: Thanks Thank you so much for the solution to my problem--it's great! Warm regards, Roger |
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