Q: For how many integer values of x is the value of 4000(2/5)^x an integer?

A: 9

This can be done with guess and check either by hand or with a calculator.

If x = 0, then 4000(2/5)^0 = 4000 so 0 works.

If x = 1, then 4000(2/5)^1 = 1600 (4000 x 0.4) so 1 works.

If x = 0, then 4000(2/5)^0 = 4000 so 0 works.

If x = 1, then 4000(2/5)^1 = 1600 (4000 x 0.4) so 1 works.

If x = 2, then 4000(2/5)^2 = 640 (1600 x 0.4) so 2 works.

If x = 3, then 4000(2/5)^3 = 256 (640 x 0.4) so 3 works.

If x = 4, then 4000(2/5)^4 = 102.4 so 4

We need to also try negative integers so here goes:

If x = 3, then 4000(2/5)^3 = 256 (640 x 0.4) so 3 works.

If x = 4, then 4000(2/5)^4 = 102.4 so 4

*does not*work.We need to also try negative integers so here goes:

If x = -1, then 4000(2/5)^-1 = 10000 (4000 x 2.5) so -1 works.

If x = -2, then 4000(2/5)^-2 = 25000 (10000 x 2.5) so -2 works.

If x = -3, then 4000(2/5)^-3= 62500 (25000 x 2.5) so -3 works.

If x = -4, then 4000(2/5)^-4 = 156250 (62500 x 2.5) so -4 works.

If x = -2, then 4000(2/5)^-2 = 25000 (10000 x 2.5) so -2 works.

If x = -3, then 4000(2/5)^-3= 62500 (25000 x 2.5) so -3 works.

If x = -4, then 4000(2/5)^-4 = 156250 (62500 x 2.5) so -4 works.

If x = -5, then 4000(2/5)^-5 = 390625 (156250 x 2.5) so -5 works.

We have been ok all along because we keep getting even numbers (until x = -4) but now…

If x = -6, then 4000(2/5)^-6 = 976562.5 (390625 x 2.5) so -6 does not work.

There are 9 integers that work:

-5, -4, -3, -2, -1, 0, 1, 2, 3

-5, -4, -3, -2, -1, 0, 1, 2, 3