How many positive integers smaller than 1,000,000 are powers of 2, but are not powers of 8? You may find it useful to consider that 2^10 = 1024.

NaturalDisaster Jun 5, 2021

#1**+3 **

2^19 is the higher power of 2 lower than 1000000. 8^6 is the highest power of 8 lower than 1000000.

19-6=13

2^13

JKP1234567890 Jun 5, 2021

#2**+4 **

Because $2^{10} = 1024,$ and $1000^2 = 1,000,000,$ we can see that $(2^{10})^2 = 1024^2$ which is obviously greater than $1,000,000.$

If we have $2^{19},$ that is $2^10 \cdot 2^9 = 1024 \cdot 512$ which is obviously less than $1,000,000.$

Thus, $2^{19} < 1,000,000 < 2^{20}$

Note $8^{n} = 2^{3n}.$

Since we know that $2^{19} < 1,000,000$ and $2^{3n} < 1,000,000, n = \left \lfloor{\frac{19}{3}}\right \rfloor = 6.$

Thus, $8^6 < 1,000,000 < 8^7.$

We have the answer to be $19 - 6 = \boxed{13}$

MathProblemSolver101 Jun 5, 2021