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 mod 4 = 1 → position 1 in cycle → 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 × 11 = a (a+b) b, carrying as needed.

Example: 47 × 11 → 4 (4+7=11) 7 → handle the carry: 4 (11) 7 → becomes 517.

For three digits abc × 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:

n × (first + last) / 2

Example: sum 1..100 = 100 × (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 multiple of 11.

Example: check 2728 for divisibility by 11: (2 + 2) - (7 + 8) = 4 - 15 = -11 → 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 (which is 7^2).
• Linear approx: sqrt(50) ≈ 7 + (50 - 49) / (2 × 7) = 7 + 1/14 ≈ 7.071.

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

Practice set (5 minutes)

1. Last digit: find last digit of 3^47.
2. Multiply 253 by 11 quickly.
3. Sum of first 75 odd numbers.
4. Is 123456 divisible by 3? By 9?
5. Approximate sqrt(200).

Answer key (brief):

1. 3^47 → cycle length 4 → 47 mod 4 = 3 → last digit 7.
2. 253 × 11 → 2 (2+5=7) (5+3=8) 3 → 2783.
3. First 75 odd numbers sum = 75^2 = 5625 (since sum of first n odd numbers = n^2).
4. 1+2+3+4+5+6 = 21 → divisible by 3 only.
5. sqrt(200) ≈ sqrt(196) + (200-196)/(2×14) = 14 + 4/28 ≈ 14.142.

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If you want, I can add a printable one-page cheat sheet from this post for quick revision before a test.