Math Core

Many MATHCOUNTS Sprint questions and early AMC 8 problems look like long, ugly calculations. They almost never are. The numbers are chosen so that a pattern, a factoring trick or a cancellation does the work for you. The skill is to look for structure before you compute.

Regroup and pair

Addition and multiplication can be done in any order, so pick the order that makes round numbers. For $4 \times 37 \times 25$ , multiply $4 \times 25 = 100$ first: the answer is $3700$ .

With long sums, pair terms that combine nicely. In an alternating sum like $2 - 4 + 6 - 8 + \dots$ , each pair of neighbors gives the same result, so you only need to count the pairs.

A useful fact about pairing: the sum of the first $n$ odd numbers is $n^2$ . For example, $1 + 3 + 5 + 7 = 16 = 4^2$ .

Factor out what's common

The distributive property works backwards too: $ab + ac = a(b + c)$ . So

$37 \times 99 + 37 = 37 \times 99 + 37 \times 1 = 37 \times 100 = 3700.$

Whenever the same number appears in several products, pull it out.

Difference of squares

$a^2 - b^2 = (a - b)(a + b)$

Read backwards, it turns a product of two numbers that are equally far from a round number into a subtraction:

$(n - k)(n + k) = n^2 - k^2.$

For example, $47 \times 53 = (50 - 3)(50 + 3) = 2500 - 9 = 2491$ .

Replace a big number by a letter

When the same huge number (or its neighbors) keeps appearing, call it $n$ . Then algebra shows what cancels. For $2026^2 - 2025 \times 2027$ , let $n = 2026$ :

$n^2 - (n - 1)(n + 1) = n^2 - (n^2 - 1) = 1.$

Telescoping

A telescoping product or sum is one where almost every piece cancels with its neighbor, leaving only the ends. Two patterns show up constantly:

$\frac{2}{3} \cdot \frac{3}{4} \cdot \frac{4}{5} \cdots \frac{9}{10} = \frac{2}{10}$

(each numerator cancels the previous denominator), and

$\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1},$

which turns a sum of fractions into a chain of cancelling differences.

Common mistake

Don't start multiplying just because the problem is written as a product. If you catch yourself computing a four-digit times four-digit product on a MATHCOUNTS Sprint round, stop and look for the trick. There almost always is one.

Worked example: An alternating sum

Compute $2 - 4 + 6 - 8 + 10 - 12 + \dots + 98 - 100$ .

Group into pairs: $(2 - 4) + (6 - 8) + \dots + (98 - 100)$ . Each pair is $-2$ . The positive terms are $2, 6, 10, \dots, 98$ , which is $25$ numbers, so there are $25$ pairs. The sum is $25 \times (-2) = -50$ .

Worked example: Difference of squares

Compute $73^2 - 27^2$ .

$73^2 - 27^2 = (73 - 27)(73 + 27) = 46 \times 100 = 4600.$

Worked example: Name the big number

Compute $\dfrac{2026^2}{2025 \times 2027 + 1}$ .

Let $n = 2026$ . The denominator is $(n - 1)(n + 1) + 1 = n^2 - 1 + 1 = n^2$ . So the fraction is $\dfrac{n^2}{n^2} = 1$ .

Worked example: A telescoping product

Compute $\left(1 - \dfrac{1}{2}\right)\left(1 - \dfrac{1}{3}\right)\left(1 - \dfrac{1}{4}\right) \cdots \left(1 - \dfrac{1}{50}\right)$ .

Each factor simplifies: $1 - \dfrac{1}{k} = \dfrac{k - 1}{k}$ . So the product is

$\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdots \frac{49}{50}.$

Every numerator from $2$ to $49$ cancels the denominator before it. Only the first numerator and the last denominator survive: $\dfrac{1}{50}$ .

Tip

Check a trick on a tiny case first. For example, $\dfrac{1}{2} \cdot \dfrac{2}{3} \cdot \dfrac{3}{4} = \dfrac{1}{4}$ confirms that only the ends survive.