Word Problem (Computer Error) Help, PLEASE?

Question

A computer was printing a sequential list of positive integers, but because of a glitch in the software, it neglected to print any numbers that were integral powers of integers (the powers were larger than 1). Thus, the list began 2, 3, 5, 6, 7, 10, 11, 12, 13, 14, 15, 17,....How many numbers in this list had fewer than four digits?

Please explain each step!

Thanks!

Answer 1

The answer is 959.*
There are 999 numbers : 0>n<1000
we only need to check 31 numbers because 32^2 >999

31 perfect squares are not printed
also 6 cubes, 1^3,4^3, and 9^3 are duplicates
no other even powers since if n is even a^n=(a^n/2)^2 and that is a duplicate

2 fifth powers (32 and 243)
only 2^7 counts

Other integral powers are duplicates
So a total of 40 numbers aren't printed

only 999 - 40 = 959 are printed

*Confirmed using Excel and automatically removing duplicates

Answer 2

The integers in the given range are 2-999
The squares are 4, 9, 16, 25, 36, ... up to 961 (31*31) or 30
The cubes are 27, 64, 125, 36*6=216, 49*7=343, 64*8=512, 9*81=729, (10 cubed is too big) or 7
The powers of 4 are 64, 81, 256 and 625 (2,3,4,5) or 4
The powers of 5 are 32, 243 (2,3) or 2
So the first idea would be 30+7+4+2 from 998 or 955 but there is overlap. 32 appears in powers of 2 & 5, 64 in 2, 3 & 4, 81 in 2&4, 256 in 2 &4, 625 in 2&4 (which takes out all in powers of 4), which increases the ones left by 6 or 961
I may have missed some duplicates

Answer 3

First lets find out how many of these powers are less than 1000.
2^2=4 and 2^9=512 for a total of eight powers of 2
3^2=9 and 3^6=729 for a total of five powers of 3
skip powers of 4 since these are powers of 2
5^2=25 and 5^4=625 for a total of three powers of 5
we continue with two powers of 6, two powers of 7 and one power of 10,11,12...31..here a total of twenty two powers
These add to 42 powers total, and excluding 1 we exclude a total of 43 numbers from the 999 numbers less than 1000
for a final total of 956 numbers..you might check this to make sure i counted correctly. Actually there are some numbers counted twice:squares of 16,25,36 which are added back in
to get 959..but check it anyway

Answer 4

ans 964
solution:
1000 is th smallest number that has 4 digits, and its square root is between 31 and 32.
So you want to discount certain powers of the follwing number, noting that
3rt(1000) = 10
4rt(1000) < 6
5rt<4
6rt<4
7rt<3
2^9<1000.
We drop
for squares, 31 - 11+1= 21
3s, [10 - 6 + 1]x2(because of their squares] = 10
3 multiple-powers of 4 are less than 1000
3 multiple-powers of 5 are less than 1000
8 multiple-pwers of 2.
1 unique power of 1
999 - 1 - 8 -3-3-10-21= 953
We have unfairly deducted 3 powers of 4, 2 of 8, 1 of 16, 1 of 32, 2 of 9, 1 of 27and 1 of 25, so the actual answer is
953 +3+2+1+1+2+1+1= 964