SOLUTION TO PROBLEM 12

If x,y and z are numbers whose base 10 representations are (abc), (bca) and (cab), respectively, then one can easily verify that

$$10x-y=999a,\qquad 10y-z=999b,\qquad 10z-x=999c.$$

Since 999 is a multiple of 37, it follows from the equations above, that if any one of x,y or z is divisible by 37, then so are the others. Consequently, one may restrict attention to multiples of 37 of the desired form. These are: 37000, 37037, 37074, 37111, $\ldots$, 37999. Since $999=27\cdot 37$, there are 28 of them.

Questions and/or comments should be directed to Judy Downey or Griff Elder

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Last modified:   Fri Nov 16 16:52:21 CST 2001