Factors & Multiples - Pre Algebra

1. If $x$ and $y$ are positive integers such that $x$ divides $y^2$ and $y$ divides $x^3$, and $\gcd(x,y) = 1$, what can be concluded about $x$ and $y$?
   Answer: Both $x$ and $y$ must equal 1 Since $\gcd(x,y) = 1$, $x$ and $y$ share no common prime factors. If $x$ divides $y^2$, then every prime power in the factorization of $x$ must appear in $y^2$ with at least the same exponent. But since $\gcd(x,y) = 1$, $x$ and $y$ have no common prime factors, so $x$ can only divide $y^2$ if $x = 1$. Similarly, since $y$ divides $x^3$ and $\gcd(x,y) = 1$, we must have $y = 1$. Therefore, both $x = 1$ and $y = 1$.
2. A number $N$ has the property that when divided by 12, the remainder is 8, and when divided by 18, the remainder is 14. What is the remainder when $N$ is divided by the least common multiple of 12 and 18?
   Answer: $32$ We have $N \equiv 8 \pmod{12}$ and $N \equiv 14 \pmod{18}$. Notice that $N = 12q + 8$ and $N = 18r + 14$ for some integers $q$ and $r$. Also, $N + 4 \equiv 0 \pmod{12}$ and $N + 4 \equiv 0 \pmod{18}$, so $N + 4$ is a multiple of both 12 and 18. The LCM of 12 and 18 is $\text{lcm}(12,18) = \frac{12 \times 18}{\gcd(12,18)} = \frac{216}{6} = 36$. So $N + 4 \equiv 0 \pmod{36}$, which means $N \equiv -4 \equiv 32 \pmod{36}$.
3. The product of two positive integers is 360. If their greatest common factor is 6, what is the minimum possible value of their least common multiple?
   Answer: $60$ Let the two numbers be $a$ and $b$. We know $ab = 360$ and $\gcd(a,b) = 6$. Since $\gcd(a,b) \times \text{lcm}(a,b) = ab$, we have $6 \times \text{lcm}(a,b) = 360$, so $\text{lcm}(a,b) = 60$. The least common multiple is always the same value (60) regardless of which specific pair $(a,b)$ we choose, as long as $ab = 360$ and $\gcd(a,b) = 6$.
4. How many positive integers less than 100 are multiples of 6 but not multiples of 9?
   Answer: $11$ Multiples of 6 less than 100: $6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90, 96$. That's $\lfloor 99/6 \rfloor = 16$ multiples. Among these, we need to exclude those that are also multiples of 9. A number is a multiple of both 6 and 9 if and only if it's a multiple of $\text{lcm}(6,9) = 18$. Multiples of 18 less than 100: $18, 36, 54, 72, 90$. That's $\lfloor 99/18 \rfloor = 5$ multiples. Therefore, there are $16 - 5 = 11$ positive integers less than 100 that are multiples of 6 but not multiples of 9.
5. The least common multiple of two numbers is 180, and their greatest common factor is 12. If one of the numbers is 36, what is the other number?
   Answer: $60$ For any two numbers $a$ and $b$, we have $\gcd(a,b) \times \text{lcm}(a,b) = a \times b$. Given that $\gcd = 12$, $\text{lcm} = 180$, and one number is 36, we have: $12 \times 180 = 36 \times b$, so $2160 = 36b$, which gives $b = 60$. We can verify: $\gcd(36, 60) = 12$ and $\text{lcm}(36, 60) = 180$.

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