Math Question, Help Please!?

Question

A "perfect power" is a number that can be written as a positive integer raised to an integer power greater then one. For example, 125 is a perfect power because it is equal to 5 cubed. The list 2, 3, 5, 6, 7, 10, 11, 12, 13, 14, 15, 17, ... contains every positive integer less than 1000 that is not a perfect power. How many integers are in the list?

Answer 1

we need to elminate perfect power.

power 2 there are 30 terms 2^2 to 31^2 (32^2 = 1024 > 1000)
power 3 there are 9 terms 2^3 to 10^3
power 4 is considered in power 2
power 5 there are 2 term 2^5 and 3^5
power 6 considered in power 2
power 7 one term 2^7
power 8 and 9 considered
1 is excluded in both list( ONE AND NOT POWER 1)
So number of terms remain = 999-(30+9+2+1) or 957

Answer 2

We start with 1000 and remove those which are perfect powers.

Let's first determine those which are cubes, which are less than or equal to 1000, which would total 31:

1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441, 484, 529, 576, 625, 676, 729, 784, 841, 900, 961

Now for those raised to the third power, which are not included above (total of 7 more):

8, 27, 125, 216, 343, 512, 1000

Those raised to the fourth power are already included in those raised to the second power, so no more here.

Those raised to the fifth power, which are not included above (2 more):

32, 243

Those raised to the sixth power are already included above.

Those raised to the seventh power not included above exceed 1000, so we know we're done.

Now count them up:

1000 - 31 - 7 - 2 = 960

960 integers are not perfect powers between 1 - 1000.

Answer 3

I don't think there is any neat trick to solve this problem, just work out a logical way to count them.

You start with 998 numbers, 2 through 999. There are nine powers of 2 (up to 2^9) that are less than 1000, so exclude 8 numbers. You have 990 left.

The powers of 3 exceed 1000 after 3^6, so exclude 5 more. You have 985 left.

Skip 4, because that is a power of 2.

The powers of 5 exceed 1000 after 5^4, so exclude 3 more. You have 982 left.

The powers of 6 exceed 1000 after 6^3, so exclude 2 more. You have 980 left.

The powers of 7 exceed 1000 after 7^3, so exclude 2 more. You have 978 left.

Skip 8 (2^3) and 9 (3^2).

Exclude 10^2. You have 977 left.

Now, for the remaining powers, only perfect squares need to be considered, since the cubes exceed 1000. Also, you can stop after 31, since 32^2 exceeds 1000. So, exclude one more number for the squares of 11, 12, 13, 14, 15, 17, 18, 19, 20, 21, 22, 23, 24, 26, 28, 29, 30, and 31. That's 18 more numbers, bringing the final total to 959.

* * * *

Note to math_kp: Your approach counts two numbers twice - 64 (4^3 and 8^2) and 729 (9^3 and 27^2)

...and note to sep_n: You missed 2^7 = 128

Answer 4

Find all the ones that ARE, count them, then subtract from 1000 to see how many aren't.

2^9 is 512 so there are 9-1 = 8 since 2^10 goes over 1000 and you can't use 1
3^6 is 729 so there are 6-1 = 5 more

skip 4 because they are included in the twos

5^4 is 625 so 4-1 = 3 more etc

be sure to skip all powers of previously done numbers (8, 9, 16, 25, 27) You can stop at 31 because 32^2 is over 1000 (and a power of 2)

Answer 5

There are 999 positive integers less than 1000.
It's easier to figure how many perfect powers there are and then subtract.
Powers of 2 = 4, 8 , 16, 32, 64, 128, 256, 512
Powers of 3 = 9, 27, 81, 243, 729
Powers of 4 (nope, those are included in powers of 2)
Powers of 5 = 25, 125, 625
Powers of 6 = 36, 216
Powers of 7 = 49, 329
Powers of 8 (nope, those are included in powers of 2)
Powers of 9 (nope, those are included in powers of 3)
Powers of 10 = 100
Each number from 11 thru 31 has one square that's less than 1000 except 16 (included in powers of 2), 25 (included in 5), and
27 = (included in 3)

Now, it's just a matter of counting

Answer 6

12

Answer 7

err.......probaly...... 46...not sure