Sam writes down positive integers, without squares and cubes. Its starts as \(2, 3, 5, 6, 7, 10, 11, ...\), what is the 1000th term, or number

Lilliam0216 Jun 23, 2024

#1**+1 **

so i would start this problem by starting with a number like 1300 then getting rid of the squares and cubes less that 1300 and see how close i am. so 1300 is about 1296 or 36^2. 10^3 is the greatest cube less than 1300. so there would be 36+10 or 46 squares and cubes. but we overcounted the 6th powers. 1, 2 and 3 are in the range so we must substract 3. 43 in total. which means our initial guess is way too high. so ill try 1050. there are still 10 cubes but only 32 squares. so there are 39 values that work. so still too high. ill try 1030. still 10 cubes and 32 squares. now ill try \(\boxed{1039}\), and i think it works!!!

shmewy Jun 23, 2024

#2**+1 **

In the first 1000 numbers, there are \(31\) perfect squares, \(10\) perfect squares, and 3 of both. Thus there are

\(1000-(31+10) + 3 = 1000 - 38 = 962 \)

So we have 38 numbers to go!

We have \(32^2=1024\), so before that we tack on another 23 numbers. We have 15 more numbers.

The next number we worry about is \(33^2=1089\)

After 1024, the next 15th number is 1039.

So 1039 is our answer.

I might be wrong though...

Thanks! :)

NotThatSmart Jun 23, 2024