suppose that a and b are positive integers both divisible by 19

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suppose that a and b are positive integers both divisible by 19. is it possible to find intergers x and y such that 0<ax+by<19? Why or why not?

Oct 1st, 2015

Here is what you are given:

a and b are positive integers that are both divisible by 19.

x and y must also be integers.

The smallest number possible for a and b is 19. Suppose both a and b are 19.

Then you would need to find integers for x and y so that 0<19x+19y<19

If both x and y are positive integers, then all sums produced by adding 19x and 19y would be greater than 19.

If both x and y are negative integers, then all sums produced by adding 19x and 19y would be less than 0.

So we are left to assume that either x or y is negative, and the other variable is positive. It does not matter if x or y is negative, it only matters that only one of them is negative.

Case where absolute values of x and y are equal:

If x = -1 and y = 1, then we would have the following:

0 < 19(-1)+19(1) <19 which is equivalent to 0 < 0 < 19, which is not true.

This case shows why the absolute values of x and y cannot be equal. Even if the absolute values of x and y are greater than 1 (while still being equal to each other), we would end up with the same result.

Case where absolute value of y is greater than the absolute value of x:

If x = -1 and y = 2, then we would have the following:

0 < 19(-1)+19(2) <19 which is equivalent to 0 < 19 < 19, which is not true.

The case gets worse if the absolute value of y becomes increasingly greater than the absolute value of x:

If x = -1 and y = 3, then we would have the following:

0 < 19(-1)+19(3) <19 which is equivalent to 0 < 38 < 19, which is not true.

The case is similar in the opposite direction, where the absolute value of x becomes increasingly greater than the absolute value of y:

Case where absolute value of y is greater than the absolute value of x:

If x = -2 and y = 1, then we would have the following:

0 < 19(-2)+19(1) <19 which is equivalent to 0 < -19 < 19, which is not true.

If x = -3 and y = 1, then we would have the following:

0 < 19(-3)+19(1) <19 which is equivalent to 0 < -38 < 19, which is not true.

Showing these three cases displays the logic of why it is NOT possible to  find integers x and y such that 0<ax+by<19 .

If your class is not a proofs class, then this is enough of an explanation to show why it is not possible.

I hope this helps you move forward on the problem! Let me know if you need any more clarification!
Oct 1st, 2015

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Oct 1st, 2015
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Oct 1st, 2015
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