5 Quick Math Tricks Every Competitive Exam Taker Should Know

Competitive exams reward a combination of accuracy and time management. In this post I’ll walk through five small, high-ROI math tricks that are easy to learn, reliable under pressure, and repeatedly useful across arithmetic, number theory and estimation problems.

If you practice these for a few days with short drills, they can save you multiple minutes during a full-length test — often the difference between a correct answer and running out of time.

1 — Last digit of powers (fast cycles)

Many contest problems only need the last digit of a power. Digits 0–9 have short repeating cycles when raised to powers. Memorize the cycles for 2, 3, 4, 7, 8, 9 and use mod arithmetic.

Example: last digit of $2^{53}$ .

Cycle for 2: 2, 4, 8, 6 (length 4).
Compute $53 \pmod 4 = 1 \implies$ position 1 in cycle $\implies$ last digit = 2.

Practice tip: write the cycle once and practice a few examples ( $3^{14}$ , $7^{23}$ ) until the mapping becomes reflexive.

2 — Multiply by 11 quickly (mental trick)

This is a reliable trick for two- and three-digit numbers.

For a two-digit number $ab$ (digits $a$ and $b$ ): $ab \times 11 = a (a+b) b$ , carrying as needed.

Example: $47 \times 11 \to 4 (4+7=11) 7 \to$ handle the carry: $4 (11) 7 \to 517$ .

For three digits $abc \times 11$ , do: $a (a+b) (b+c) c$ and carry from right to left.

Why it helps: it turns a multiplication into 2–3 quick mental additions instead of full multiplication.

3 — Sum of arithmetic series (closure trick)

If numbers form an arithmetic progression, use the formula:

$\frac{n \times (\text{first} + \text{last})}{2}$

Example: sum $1 \dots 100 = \frac{100 \times (1 + 100)}{2} = 5050$ .

Practical use: recognizing an arithmetic progression in a problem removes need for long addition and avoids mistakes under time pressure.

4 — Divisibility shortcuts (digit-sum & alternating sum)

These are quick checks to filter options or identify divisible numbers without division.

Divisible by 3: sum of digits divisible by 3.
Divisible by 9: sum of digits divisible by 9.
Divisible by 11: alternating digit-sum rule: (sum of digits at odd positions) - (sum at even positions) is a multiple of 11.

Example: check 2,728 for divisibility by 11: $(2 + 2) - (7 + 8) = 4 - 15 = -11 \implies$ divisible by 11.

5 — Quick square-root approximations

When exact square roots are heavy, approximate using the nearest perfect square and linearize.

Example: $\sqrt{50}$

Nearest perfect square: 49 ( $7^2$ ).
Linear approx: $\sqrt{50} \approx 7 + \frac{50 - 49}{2 \times 7} = 7 + \frac{1}{14} \approx 7.071$ .

This is particularly useful for estimation problems and eliminating answer choices in multiple-choice settings.

Practice set (5 minutes)

Last digit: find last digit of $3^{47}$ .
Multiply 253 by 11 quickly.
Sum of first 75 odd numbers.
Is 123,456 divisible by 3? By 9?
Approximate $\sqrt{200}$ .

Answer key

$3^{47} \implies$ cycle length 4 $\implies 47 \pmod 4 = 3 \implies$ last digit 7.
$253 \times 11 \implies 2\,(2+5=7)\,(5+3=8)\,3 \implies 2,783$ .
First 75 odd numbers sum $= 75^2 = 5,625$ (since sum of first $n$ odd numbers $= n^2$ ).
$1+2+3+4+5+6 = 21 \implies$ divisible by 3, but not by 9.
$\sqrt{200} \approx \sqrt{196} + \frac{200-196}{2 \times 14} = 14 + \frac{4}{28} \approx 14.142$ .

Frequently Asked Questions (FAQ)

What is the cycle length for exponents of other base digits?

Base digits 0, 1, 5, 6 always end with themselves (cycle length 1). Digits 4 and 9 alternate with period 2 ( $4^1=4, 4^2=16$ ). Digits 2, 3, 7, and 8 repeat in cycles of 4.

Can linear approximation be applied to cube roots?

Yes. Using differential approximation, $\sqrt[3]{x + \Delta x} \approx \sqrt[3]{x} + \frac{\Delta x}{3 \cdot (\sqrt[3]{x})^2}$ . For example, $\sqrt[3]{28} \approx 3 + \frac{1}{3 \times 9} = 3 + \frac{1}{27} \approx 3.037$ .

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