🚀 go-pugleaf

On Fri, 02 May 1997 14:32:16 -0600, srinivasan@pspro.org.kw wrote:

>Articles X,Y and Z are for sale.Article X can be bought at the rate of
>eight for $1, Article Y costs $1 each and article Z costs $10 each.I buy
>a selection of all  three types and find that I have purchased exactly
>100 articles at a cost of $100 .How many articles of each type do I have?

Here's a solution without trial-and-error.  Let x, y, and z be the
number of articles X, Y, and Z, respectively, that you've purchased.
Then x, y, and z must be integers, 

x + y + z = 100, and
x/8 + y + 10z = 100.

Equating the left-hand sides and simplifying yields

x + z = x/8 + 10z, 
7x/8 = 9z, or
7x = 72z.

Now x is an integer, so 7 is a divisor of 72x.  Since 7 is a prime
number, and 7 is not a divisor of 72, this implies that 7 is a divisor
of z.  Similarly, 72 is a divisor of 7x; and since 72 and 7 have no
common factor bigger than 1, 72 must be a divisor of x.  Clearly x
cannot be greater than 100, so either x = 0, or x = 72.  However,
since you bought a selection of *all* three types, x cannot be 0, and
therefore x = 72.

If x = 72, then 7 * 72 = 72z, so z = 7, and y = 100 - x - z = 21.  Is
this a solution?  x/8 + y + 10z = 9 + 21 + 70 = 100, so it is.

Brian M. Scott