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Math Challenge 1-A Number Theory Problem 3.26 and 3.20

Re: Math Challenge 1-A Number Theory Problem 3.26 and 3.20

For both of these problems it's important to think in terms of prime factorizations.

Here's some thoughts:

Problem 3.26: We need something to be a perfect square and a perfect cube. For something to be a perfect square, all the exponents in the prime factorization need to be even (for example $36 = 2^2\cdot 3^2$) and for something to be a perfect cube, all the exponents in the prime factorization need to be a multiple of 3 (for example, $1000 = 2^3\cdot 5^3$).

For both of these to be true, we need all the exponents to be a multiple of $6$. Since our number is $200\cdot x = 2^3\cdot 5^2\cdot x$, how many extra $2$s and extra $5$s do we need? That allows us to find $x$.

Problem 3.30 is similar to Example 3.10, so you can look at that answer in the book as well. But let's think about a few examples to start.

If we remove all the multiples of $2$ except $2$, we are left with$$2, 3, 5, 7, 9, 11, 13, \ldots$$Are all the composite numbers removed? Not yet, and the smallest that is still there is $9 = 3^2$.

So, continuing, we next remove all the multiples of $3$ except $3$, leaving us with$$2, 3, 5, 7, 11, 13, 17, 19, 23, 25, 29, \ldots$$We're not done yet, as composite numbers still remain. Notice, however, that the smallest left now is $25 = 5^2$.

So, now try removing all the multiples of $5$ except $5$ and see what remains. Hopefully you start to see a pattern here and that helps answering the full question. Hint: You don't have to remove that many primes, so if you're still stuck, try to just keep going until you remove all the composite numbers!