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The 22 Puzzle


Date: 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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