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Anthony

For the first problem, I got z=1, x=2, y=1. z=1, x=1, and y=1 would work if two variables are allowed to be the same. If you solve over the reals, then you could use zero and negatives, and their would be a whole lot more correct answers, like x=9, y=-1 and z=0.

For the second problem, I got 30 for three distinct prime factors and 210 for four distinct prime factors.

For the third problem, I got 50 k’s as if k is even, it will work.

For the fourth problem, I a remainder of 25. If you cleverly add it (1+24 + 2+23…) you get 25*12. Since there is a 25, the remainder is zero. Another way to do it is only looking at the ones place while adding. You get 1+2+3+4+5 = 5 if you only look at the ones digit. Then just add on and use the divisibility rule for 5.

For the fifth problem, I got 37 years old.