Question 240770
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You only missed it by THAT much (holds thumb and forefinger a tiny distance apart)


Your method would have worked just fine if he had only two tickets (of course, in that case, the sum would necessarily have been odd).  But he had three tickets, so you need three consecutive integers.


If you let *[tex \Large x] represent the first integer, then *[tex \Large x\ +\ 1] is most certainly the next consecutive integer -- just as you stated.  But you needed the next one as well, which is *[tex \Large (x\ +\ 1)\ +\ 1\ =\ x\ +\ 2]


Now, the sum of all <i>three</i> is 138, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ (x\ +\ 1)\ +\ (x\ +\ 2)\ =\ 138]


Solve that and you will find that *[tex \LARGE x] is a nice, tidy integer.


Write back and let me know how you did.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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