The 22 PuzzleDate: 03/12/2002 at 06:59:44 From: sam hughes Subject: The 22 puzzle 1. Choose three different digits from 1-9. 2. Make all the two-digit numbers you can from these three digits, then add them. 3. Add the three original digits and divide into the sum from step 2. E.g 1,6,8. Make the numbers 16,18,61,81,68,86 = 330 1+6+8 = 15 and 330 divided by 15 = 22. Why is the answer always 22? Kind regards, Sam Date: 03/12/2002 at 11:58:33 From: Doctor Pete Subject: Re: The 22 puzzle Hi Sam, This is a fun question! The key is to express the various sums in the problem as a function of the digits you choose. So, let a, b, and c represent the three digits you choose. Then, if you pick any two digits, say a and b, and form the two-digit number corresponding to them, it would be 10a + b. For example, if a = 1, b = 6, then the result of joining these two digits is the number 16, or equivalently, 10(1) + 6. This is because in our decimal number system, the digit 1 is in the tens place, and thus represents 1 set of 10. (So 8402 = 8000 + 400 + 0 + 2.) Now, there are exactly six ways to choose two digits out of a set of three distinct digits; they are: {a,b}, {b,a} {a,c}, {c,a} {b,c}, {c,b} These correspond to the numbers 10a + b, 10b + a 10a + c, 10c + a 10b + c, 10c + b Their sum is therefore 11a + 11b + 11a + 11c + 11b + 11c = 22(a+b+c). And since the sum of the three digits we originally picked is simply a+b+c, it follows that dividing the first sum by the second gives 22(a+b+c)/(a+b+c) = 22, thus proving the answer is independent of the choice of digits. The proof of this claim may inspire you to generalize the problem with respect to the number of digits. For example, if you were to choose four distinct digits, and then form all the 3-digit numbers from these, add them up, and divide by the sum of the four digits, what would you get? Well, each three-digit number has the form 100a + 10b + c, and you can see that the various permutations of the digits are {a,b,c}, {a,c,b}, {b,a,c}, {b,c,a}, {c,a,b}, {c,b,a} {a,b,d}, etc. {a,c,d}, etc. {b,c,d}, etc. I didn't need to fill in rest because all I needed to know is that each row has six triples, and in each row the sum of the resulting three-digit numbers is going to be 222(a+b+c), 222(a+b+d), 222(a+c+d), and 222(b+c+d). Then the whole sum is 666(a+b+c+d), and dividing by the sum of the original four digits gives 666. I suppose if you are superstitious this result may give you pause, but to a mathematician it's just a number. Finally, if we took the same number of digits (four), but paired them instead of putting them in triplets, the result would be 33. Can you prove it? - Doctor Pete, The Math Forum http://mathforum.org/dr.math/ Date: 03/12/2002 at 12:05:33 From: Doctor Ian Subject: Re: The 22 puzzle Hi Sam, It's easier to see what's happening if we give the digits names, instead of using actual digits. Suppose the digits we choose are a, b, and c. Then the numbers we can make from them are ab, ac, ba, bc, ca, cb Now to try to add these together would be a mess, because we'd have to worry about possible carries. But we can use a trick. Well, not a trick, really, but the definition of what it _means_ to write a number like 'ab': ab = 10*a + b Make sure you understand this before going on. If it doesn't make sense, write back and I'll explain it in more detail. Anyway, now we can add the numbers up easily: 10a + b 10a + c 10b + a 10b + c 10c + a 10c + b -------------------------- 10(2a+2b+2c) + (2a+2b+2c) So the sum will always be (10 + 1)(2a + 2b + 2c) = 11(2a + 2b + 2c) = 22(a + b + c) Note that this is true no matter _how_ we choose a, b, and c. So, when you add up the digits and divide, what would you expect to get for a result? If you play with this a little, you can see that if you make all the three-digit numbers, abc, acb, bac, bca, cab, cba add them up, and divide by the sum of the digits, you'll get 222 instead of 22. There is a whole range of tricks that you could make up this way, by altering the number of digits, and the sizes of the numbers you make with them. Does this help? - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ |
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