Number Properties

What you need to know

• Parity (odd and even). Odd $\pm$ odd $=$ even, and odd $\times$ odd $=$ odd. A sum of integers is odd exactly when it has an odd number of odd terms. A number and its square are both odd or both even: $n^2 \equiv n \pmod 2$. Example: $3+5+7+10$ has three odd terms, so it is odd ($25$).
• Perfect squares end in $0$, $1$, $4$, $5$, $6$, or $9$, never $2$, $3$, $7$, or $8$. Divided by $4$, a square leaves remainder $0$ or $1$; divided by $3$, also $0$ or $1$. An odd square leaves remainder $1$ when divided by $8$. Example: $2027$ ends in $7$, so it is not a square; $9^2 = 81 = 8\cdot10+1$.
• Perfect cubes leave remainder $0$, $1$, or $8$ when divided by $9$. Since $8 \equiv -1$, we write this as $\equiv 0, \pm1 \pmod 9$. Example: $4^3 = 64 = 9\cdot7+1$ and $5^3 = 125 = 9\cdot13+8$.
• Consecutive integers. Among any $k$ consecutive integers, exactly one is divisible by $k$. Even more: the product of any $k$ consecutive integers is divisible by $k! = 1\cdot2\cdots k$. Why: for positive integers, the product divided by $k!$ is a binomial coefficient, which counts something, so it is a whole number. Example: $5\cdot6\cdot7\cdot8 = 1680$, and $1680 \div 4! = 70 = \binom84$.
• Floor function. $\lfloor x \rfloor$ (the floor of $x$) is the greatest integer less than or equal to $x$: round down. So $\lfloor x \rfloor \le x < \lfloor x \rfloor+1$. Example: $\lfloor 3.7 \rfloor = 3$ and $\lfloor -1.2 \rfloor = -2$.
• Counting squares. There are $\lfloor \sqrt N \rfloor$ perfect squares from $1$ to $N$. Example: $\lfloor \sqrt{200} \rfloor = 14$, since $14^2 = 196$ and $15^2 = 225$.
• $n^2-1 = (n-1)(n+1)$. If $n$ is odd, this is divisible by $8$. If $3$ does not divide $n$, it is divisible by $3$. Example: $11^2-1 = 120 = 8\cdot15 = 3\cdot40$.
• Two sums. $1+2+\dots+n = \frac{n(n+1)}{2}$, and the first $n$ odd numbers add to $n^2$. Example: $1+2+\dots+20 = 210$ and $1+3+5+7+9 = 25$.

How AMC 10 tests it

• "How many integers $n$ in a range make $n^2-1$ (or $n^3-n$) divisible by $24$?" Parity and remainders (problems 10–16).
• Counting squares or cubes in a range, or numbers that are neither: floors of square and cube roots plus the Principle of Inclusion-Exclusion (PIE).
• Parity arguments: a sum of $2025$ odd numbers can't be $2024$; chessboard-coloring arguments that something is impossible.
• Sequences of integers with a growth rule: find how many terms are below $N$, or a particular term, using floors.
• "Which of the following must be true about $n$?" with five statements about odd and even, squares, and products of consecutive numbers.

Standard approaches

1. Check parity first; it rules out half the possibilities in seconds.
2. Factor the expression, then check what divides each factor. Example: $n(n+1)$ is always even, since one of two consecutive integers is even.
3. Work mod $3$, $4$, $8$, or $9$, where squares and cubes have few possible remainders.
4. To count from $1$ to $N$, use floors and PIE, and double-check the endpoints.
5. When a statement "must" hold, test $n = 1$, $n = 2$, and a prime. One counterexample settles it.

Worked example

For how many integers $n$ with $1 \le n \le 100$ is $n^2 - 1$ divisible by $24$?

(A) $25$ (B) $30$ (C) $33$ (D) $34$ (E) $50$

Step 1: split $24$. $n^2-1 = (n-1)(n+1)$ and $24 = 8\cdot3$. Since $8$ and $3$ share no factor, check each one separately.

Step 2: the $8$. If $n$ is odd, $n-1$ and $n+1$ are two even numbers in a row, like $10$ and $12$. One is a multiple of $4$ and the other a multiple of $2$, so their product is a multiple of $8$. If $n$ is even, the product is odd.

Step 3: the $3$. Exactly one of $n-1$, $n$, $n+1$ is a multiple of $3$. If $3$ does not divide $n$, it divides $n-1$ or $n+1$. If $3$ divides $n$, it divides neither.

Step 4: the condition. So $24$ divides $n^2-1$ exactly when $n$ is odd and not a multiple of $3$, that is, when $\gcd(n, 6) = 1$.

Step 5: count with PIE. From $1$ to $100$ there are $50$ even numbers, $33$ multiples of $3$, and $16$ multiples of $6$ (counted twice). That leaves $100-50-33+16 = 33$.

The answer is $\boxed{\textbf{(C)}\ 33}$.

Pitfalls

• Believing every square leaves remainder $1$ when divided by $8$. Only odd squares do. Example: $6^2 = 36$ leaves $4$.
• Miscounting squares from $a$ to $b$. The count is $\lfloor \sqrt b \rfloor - \lceil \sqrt a \rceil + 1$, where $\lceil x \rceil$ (the ceiling) means round up. Using $\lfloor \sqrt b \rfloor - \lfloor \sqrt a \rfloor$ leaves out $a$ when $a$ is a perfect square. Example: from $16$ to $50$ there are $4$ squares ($16, 25, 36, 49$): $7-4+1 = 4$, not $7-4 = 3$.
• Assuming $\lfloor x+y \rfloor = \lfloor x \rfloor + \lfloor y \rfloor$; it can be off by one. Example: $\lfloor 0.6+0.6 \rfloor = 1$, but $\lfloor 0.6 \rfloor + \lfloor 0.6 \rfloor = 0$.
• Forgetting that $n = 0$ and $n = 1$ are integers that may satisfy (or break) the condition. Example: in the worked example $n = 1$ counts, since $1^2-1 = 0$ and $24$ divides $0$.

Traps that recur

• Counting one solution for every integer n with |x| < 100 (n = -100 to 99, giving 200) without noticing that n = 99 yields only f = 1, which is excluded. (2018 AMC 10B #25)
• Counting the 45 pairs (j, k) with 1 <= j < k <= 10 as separate jump points, forgetting that equal fractions like 1/2 = 2/4 = 3/6 are one jump. (2016 AMC 10B #25)
• Forgetting the constraint r <= 2m (so that m^2 is really the largest square below n) and choosing a smaller m, or miscounting: 8 numbers in the sequence means 7 subtractions. (2010 AMC 10A #25)
• Using the naive pattern 1, 4, 7, 10, … (odds only 6 apart) or evens spaced 4 apart (506), or dropping the extra 2 elements from the leftover 2021–2024. (2024 AMC 10A #20)