Number Properties - GMAT Quantitative Reasoning

1. The positive integer $N$ has the property that when $N$ is divided by 7, the quotient is $q$ and the remainder is $r$, where $q$ and $r$ are both positive. If $q + r = 10$ and $N < 100$, how many possible values are there for $N$?
   Answer: 6 By the division algorithm, $N = 7q + r$ where $0 \leq r < 7$. Since both $q$ and $r$ are positive, we have $r \in \{1,2,3,4,5,6\}$ and $q \geq 1$. Given $q + r = 10$, we have $q = 10 - r$. Since $q \geq 1$, we need $10 - r \geq 1$, so $r \leq 9$. Combined with $1 \leq r \leq 6$, we get $r \in \{1,2,3,4,5,6\}$. The corresponding values of $q$ are $\{9,8,7,6,5,4\}$. This gives us the following values of $N$: For $r=1, q=9$: $N = 7(9) + 1 = 64$. For $r=2, q=8$: $N = 7(8) + 2 = 58$. For $r=3, q=7$: $N = 7(7) + 3 = 52$. For $r=4, q=6$: $N = 7(6) + 4 = 46$. For $r=5, q=5$: $N = 7(5) + 5 = 40$. For $r=6, q=4$: $N = 7(4) + 6 = 34$. All these values are less than 100, so there are 6 possible values for $N$.
2. If $m$ and $n$ are positive integers such that $\gcd(m,n) = 1$ and $m^2 + n^2$ is divisible by $mn$, what is the smallest possible value of $m + n$?
   Answer: 2 We need $mn | (m^2 + n^2)$ with $\gcd(m,n) = 1$. Since $\gcd(m,n) = 1$, we have $\gcd(m,mn) = m$ and $\gcd(n,mn) = n$. For $mn$ to divide $m^2 + n^2$, we need both $m | (m^2 + n^2)$ and $n | (m^2 + n^2)$. Since $m | m^2$, we need $m | n^2$. But $\gcd(m,n) = 1$ implies $\gcd(m,n^2) = 1$, so $m | n^2$ is only possible if $m = 1$. Similarly, $n | m^2$ with $\gcd(m,n) = 1$ implies $n = 1$. Therefore, $m = n = 1$, giving $m + n = 2$. Let's verify: $m^2 + n^2 = 1 + 1 = 2$ and $mn = 1$. Indeed, $1 | 2$, so this works. Also, $\gcd(1,1) = 1$ ✓. Therefore, the smallest possible value is 2.
3. If $k$ is a positive integer and the remainder when $k$ is divided by 6 is 4, which of the following is the remainder when $k^2$ is divided by 6?
   Answer: 4 When you encounter remainder problems involving powers, the key insight is that you can work with the remainder itself rather than the original large number. Since $k$ leaves remainder 4 when divided by 6, we can write $k = 6q + 4$ for some integer $q$. To find the remainder when $k^2$ is divided by 6, we only need to examine what happens when we square the remainder 4. Let's calculate: $k^2 = (6q + 4)^2 = 36q^2 + 48q + 16$ Since $36q^2$ and $48q$ are both multiples of 6, they don't affect the remainder. We only need to find the remainder when 16 is divided by 6. $16 = 6 \times 2 + 4$ Therefore, $k^2$ leaves remainder 4 when divided by 6. Looking at the wrong answers: Choice A (remainder 0) would mean $k^2$ is divisible by 6, but $4^2 = 16$ clearly isn't. Choice B (remainder 2) and Choice D (remainder 5) result from calculation errors or misconceptions about how remainders behave under multiplication. You can verify this with specific examples: if $k = 10$ (which gives remainder 4 when divided by 6), then $k^2 = 100$, and $100 = 6 \times 16 + 4$. Study tip: For remainder problems involving powers, focus on the remainder of the base number and calculate from there. This shortcut saves time and reduces errors compared to working with large numbers directly.

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