Squaring 2-Digit Numbers: Shortcuts for Endings 5, 4, 6, and 1

Squaring a 2-digit number in under ten seconds sounds like a party trick. In a timed quantitative aptitude section, it is the difference between finishing the paper and leaving questions blank.

The four patterns below cover every 2-digit number ending in 5, 4, 6, or 1, and each is derivable from a single algebraic identity. More importantly, the legacy formulas circulating online for the 4-ending and 6-ending cases contain arithmetic errors. This article re-derives all four from first principles.

Why Mental Math Speed Matters in Placement Tests

The AMCAT platform, run by SHL India, typically allows 25 to 35 minutes for a quantitative section. TCS NQT runs a similar time-pressure structure. At roughly 70 seconds per question, any calculation you shortcut from 40 seconds to 8 seconds frees up half a minute for harder problems.

Squaring shortcuts fall into a specific category of aptitude prep: one-time learning that compounds forever. Spend a single study session understanding these four patterns. Every subsequent test, you spend 8 seconds on a squaring question instead of 45.

The same time-budget pressure applies to other aptitude areas. The articles on calendar problems and clock problems cover shortcut frameworks for those categories using the same approach.

Squaring 2-Digit Numbers Ending in 5

This is the foundation. The other three shortcuts in this article are derived from it.

Rule: Take the tens digit A. Multiply A by (A + 1). Write 25 after the result.

The algebraic reason: (10A + 5)² = 100A² + 100A + 25 = 100 × A × (A+1) + 25. The “write 25 after” is literally appending 25 to 100 times the leading product.

Worked examples

• Example 1: Find 35²

  • Tens digit A = 3. Next integer = 4.
  • Step 1: 3 × 4 = 12
  • Step 2: Write 25 after: 1225
  • Answer: 1225

• Example 2: Find 75²

  • Tens digit A = 7. Next integer = 8.
  • Step 1: 7 × 8 = 56
  • Step 2: Write 25 after: 5625
  • Answer: 5625

• Example 3: Find 95²

  • Tens digit A = 9. Next integer = 10.
  • Step 1: 9 × 10 = 90
  • Step 2: Write 25 after: 9025
  • Answer: 9025

Try 45² before moving on: A = 4, 4 × 5 = 20, append 25, answer is 2025.

Squaring 2-Digit Numbers Ending in 1

For numbers ending in 1, the formula uses three components that map directly onto the algebraic expansion of (10A + 1)².

Rule: For a number with tens digit A:

• Component 1: A² × 100
• Component 2: 2 × A × 10
• Component 3: Add 1
• Answer = Component 1 + Component 2 + 1

Algebraic basis: (10A + 1)² = 100A² + 20A + 1.

Worked examples

• Example 1: Find 21²

  • A = 2
  • Step 1: 2² × 100 = 400
  • Step 2: 2 × 2 × 10 = 40
  • Step 3: 400 + 40 + 1 = 441

• Example 2: Find 71²

  • A = 7
  • Step 1: 7² × 100 = 4900
  • Step 2: 2 × 7 × 10 = 140
  • Step 3: 4900 + 140 + 1 = 5041

• Example 3: Find 91²

  • A = 9
  • Step 1: 9² × 100 = 8100
  • Step 2: 2 × 9 × 10 = 180
  • Step 3: 8100 + 180 + 1 = 8281

The pattern is consistent: the middle term (2A × 10) grows linearly, which is why 91² has a notably larger middle step than 21².

Squaring 2-Digit Numbers Ending in 4

Here the 5-ending shortcut pays off a second time. Instead of memorising a new formula, you borrow from the adjacent 5-ending square.

Rule: For any 2-digit number N4 ending in 4, its 5-ending neighbour is N5 = N4 + 1. Square N5 using the 5-trick, then subtract the sum of the two numbers.

Formula: N4² = N5² minus (N5 + N4)

Algebraic basis: (n − 1)² = n² − 2n + 1, and 2n − 1 equals n + (n − 1), so n² − (n + (n−1)) = (n−1)².

Worked examples

• Example 1: Find 24²

  • Adjacent 5-neighbour: 25. 25² = 2 × 3 followed by 25 = 625.
  • Sum of the pair: 25 + 24 = 49
  • Answer: 625 − 49 = 576

• Example 2: Find 54²

  • Adjacent 5-neighbour: 55. 55² = 5 × 6 followed by 25 = 3025.
  • Sum of the pair: 55 + 54 = 109
  • Answer: 3025 − 109 = 2916

• Example 3: Find 84²

  • Adjacent 5-neighbour: 85. 85² = 8 × 9 followed by 25 = 7225.
  • Sum of the pair: 85 + 84 = 169
  • Answer: 7225 − 169 = 7056

Note: the sum-of-pair grows by 20 for each step up the number line (49, 69, 89, 109, …), which makes it easy to sanity-check your calculation mid-problem.

Squaring 2-Digit Numbers Ending in 6

Symmetric to the 4-ending case. Add instead of subtract.

Rule: For any 2-digit number N6 ending in 6, its 5-ending neighbour is N5 = N6 − 1. Square N5 using the 5-trick, then add the sum of the two numbers.

Formula: N6² = N5² plus (N5 + N6)

Algebraic basis: (n + 1)² = n² + 2n + 1, and 2n + 1 equals n + (n+1).

Worked examples

• Example 1: Find 26²

  • Adjacent 5-neighbour: 25. 25² = 625.
  • Sum of the pair: 25 + 26 = 51
  • Answer: 625 + 51 = 676

• Example 2: Find 56²

  • Adjacent 5-neighbour: 55. 55² = 3025.
  • Sum of the pair: 55 + 56 = 111
  • Answer: 3025 + 111 = 3136

• Example 3: Find 86²

  • Adjacent 5-neighbour: 85. 85² = 7225.
  • Sum of the pair: 85 + 86 = 171
  • Answer: 7225 + 171 = 7396

The 6-ending examples also let you spot-check: 26² = 676, 36² = 1296, 46² = 2116. Each answer ends in 6, which is correct since any number ending in 6 squares to a number ending in 6.

All Four Shortcuts at a Glance

Ending	Rule	Example
5	Multiply A × (A+1), write 25 after	65² = 6×7=42, append 25, answer 4225
1	A²×100 + 2A×10 + 1	61² = 3600+120+1 = 3721
4	Adjacent_5² minus (N5 + N4)	44² = 45²−89 = 2025−89 = 1936
6	Adjacent_5² plus (N5 + N6)	46² = 45²+91 = 2025+91 = 2116

Decision tree for any squaring question in an aptitude test:

• If the number ends in 5: apply the 5-trick directly.
• If the number ends in 1: apply the three-component formula.
• If the number ends in 4: find the adjacent 5-number, apply the 5-trick, subtract.
• If the number ends in 6: find the adjacent 5-number, apply the 5-trick, add.

This decision tree takes under two seconds. The calculation takes six to eight more. Ten seconds total is faster than long-form multiplication by roughly 35 seconds.

The same pattern-based approach applies to multiplication shortcuts. The article on multiplying a number by 111 covers a related class of Vedic-style arithmetic tricks that appear alongside squaring questions in quantitative sections.

Practice Questions

Work through these without a calculator. Check your answers against the solutions below.

• Q1: Find 45²
• Q2: Find 81²
• Q3: Find 64²
• Q4: Find 76²

Solutions

• Q1: 45²

  • A = 4, 4 × 5 = 20, append 25.
  • Answer: 2025

• Q2: 81²

  • A = 8, 8²×100 = 6400, 2×8×10 = 160, plus 1.
  • Answer: 6400 + 160 + 1 = 6561

• Q3: 64²

  • Adjacent 5-neighbour: 65. 65² = 6×7=42, append 25 = 4225.
  • Sum: 65 + 64 = 129. Answer: 4225 − 129 = 4096

• Q4: 76²

  • Adjacent 5-neighbour: 75. 75² = 7×8=56, append 25 = 5625.
  • Sum: 75 + 76 = 151. Answer: 5625 + 151 = 5776

The IndiaBix numbers aptitude section has a full drill set if you want extended practice beyond these four.

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