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Classification (Odd One Out)

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high importance~3 Q in Tier 19 formulas⚡ 9 shortcuts6 subtopics
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Number odd one out

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⏱ 4 min read🧩 6 question types🎯 20 practice Q
The idea in one minute

Four numbers are given. Three share a number property and one breaks it. You do not need long calculation; you need a fixed list of properties to test in order. The list: prime or composite, perfect square or cube, multiples, digit rules, odd or even, and near-forms like a square plus one. When the first property fits all four, look one level deeper at the roots or the parity.

01

What number classification asks

Four numbers are given. Three share a number property. One does not. You need a list of properties to test in a fixed order, not long calculation.

Take 2, 4, 6, 9. Three are even and 9 is odd. So 9 is the odd one. Harder questions just use a deeper property.

02

Test properties in this order

  1. Prime or composite. A prime has exactly two factors: 1 and itself. Test a number up to 200 by dividing by 2, 3, 5, 7, 11 and 13 only.
  2. Perfect square or cube. Learn squares to 302=90030^2 = 900 and cubes to 123=172812^3 = 1728. A square never ends in 2, 3, 7 or 8.
  3. Multiples. Digit sum for 3 and 9. Alternate-digit difference for 11. Plain division for 7 and 13.
  4. Digit rules. Same digit sum or same digit product for three numbers, one different.
  5. Parity. Odd versus even. Usually the second level, when all four share another property.
  6. Number forms. Three numbers fit n2+1n^2 + 1, n3−1n^3 - 1 or n(n+1)n(n+1), and one does not.
03

Learn the fake primes

These composites look prime. Learn their factors on sight.

NumberFactors
513 × 17
573 × 19
873 × 29
917 × 13
1197 × 17
1337 × 19
14311 × 13
1617 × 23

Example: 79, 83, 89, 91. The first three have no divisor up to their square roots. But 91 = 7 × 13. Answer: 91.

04

Divisibility shortcuts

  • By 3 or 9: add the digits. 345 gives 3 + 4 + 5 = 12, so 345 divides by 3 but not 9.
  • By 11: subtract the sum of digits at even places from the sum at odd places. A result of 0 or a multiple of 11 means divisible. For 2728: (2 + 2) − (7 + 8) = −11, so 2728 divides by 11.
  • By 7 and 13: just divide. The numbers stay small.
05

Digit rules

When primes, squares and multiples all fail, look at the digits. 138, 234 and 164 all have digit product 24 (1×3×8, 2×3×4, 1×6×4). 326 has 3×2×6 = 36. So 326 is odd.

Tip: Digits can be rearranged freely while keeping the same sum and product. Shuffled digits usually hide a digit rule.

06

Two-level questions

Sometimes all four numbers pass the first test. 27, 125, 343, 512 are all cubes. Look at the roots: 3, 5 and 7 are odd; 8 is even. So 512 is the odd one.

Watch: A number can be both a square and a cube: 64 = 8² = 4³, and 729 = 27² = 9³. Check both.

07

Check your answer

A good rule makes exactly one number odd. Test one more property after you find your answer. If another natural rule points to a different number, re-read the options. In a well-made question, every natural rule points to the same answer.

08

Question types you will see

Each type: how to recognise it, the method step by step, and one question to try.

Type 1very common3 practice Q

Prime vs composite

How to spot it:

Four numbers, mostly odd, that look prime. One is a hidden product like 91 or 143 (or one prime among composites).

p prime  ⟺  no divisor 2≤d≤pp \text{ prime} \iff \text{no divisor } 2 \le d \le \sqrt{p}
Method
  1. Divide each number by 2, 3, 5, 7, 11, 13.

  2. Mark which ones have a factor.

  3. The single number on the other side is the answer.

Why it works:

Primes are the setter's favourite group, and hidden composites like 91 and 143 are the traps.

Try this

Find the odd one: 61, 67, 71, 77.

Show solution
  1. 61, 67 and 71 have no divisor up to their square roots. They are prime.

  2. 77 = 7 × 11 is composite.

  3. Rule: three primes, one composite.

Answer

77

Type 2very common2 practice Q

Perfect square / perfect cube

How to spot it:

Numbers such as 169, 196, 225 or 216, 343, 512 — familiar powers with one stranger.

n2,  n3n^2,\; n^3
Method
  1. Recall the square and cube tables.

  2. Mark which numbers are exact squares or cubes.

  3. The one that is not is odd.

Why it works:

A number between two known powers cannot itself be a power.

Try this

Find the odd one: 169, 196, 225, 250.

Show solution
  1. 169 = 13², 196 = 14², 225 = 15².

  2. 250 lies between 15² = 225 and 16² = 256.

  3. Rule: three squares, one non-square.

Answer

250

Type 3common2 practice Q

Multiples / divisibility

How to spot it:

Three numbers are multiples of one number (7, 9, 11, 13), often shown by a digit rule.

11∣n  ⟺  (odd-place sum)−(even-place sum)≡0(mod11)11 \mid n \iff (\text{odd-place sum}) - (\text{even-place sum}) \equiv 0 \pmod{11}
Method
  1. Find the common divisor of most numbers (try 7, 9, 11, 13).

  2. Use divisibility rules: digit sum for 9, alternate sum for 11.

  3. The number that is not a multiple is odd.

Why it works:

The setter picks a divisor, writes three multiples of it, and adds one near-miss.

Try this

Find the odd one: 39, 65, 91, 111.

Show solution
  1. 39 = 13 × 3, 65 = 13 × 5, 91 = 13 × 7. All are multiples of 13.

  2. 111 = 3 × 37, which is not a multiple of 13.

  3. Rule: three multiples of 13, one not.

Answer

111

Type 4common2 practice Q

Digit sum / digit product

How to spot it:

Numbers with no clear prime or square pattern, often three-digit, whose digits look shuffled.

S(n)=∑digits,  P(n)=∏digitsS(n) = \sum \text{digits},\; P(n) = \prod \text{digits}
Method
  1. Add the digits of each number.

  2. If the sums differ, multiply the digits.

  3. The number whose sum or product differs is odd.

Why it works:

Digits can be rearranged freely while keeping the same sum and product, so shuffled digits hide a digit rule.

Try this

Find the odd one: 345, 453, 534, 546.

Show solution
  1. Digit sums: 3 + 4 + 5 = 12, 4 + 5 + 3 = 12, 5 + 3 + 4 = 12.

  2. 5 + 4 + 6 = 15, which is different.

  3. Rule: three have digit sum 12, one has 15.

Answer

546

Type 5common2 practice Q

Two-level: same power, odd or even root

How to spot it:

All four are squares (or all cubes), so the first rule does not decide.

Method
  1. Write the root of each number.

  2. Check the roots for odd/even or prime/composite.

  3. The number whose root breaks the pattern is odd.

Why it works:

The first property is shared by all four to trap you; the answer hides in the roots.

Try this

Find the odd one: 27, 125, 343, 512.

Show solution
  1. All four are cubes: 3³, 5³, 7³, 8³.

  2. Roots 3, 5 and 7 are odd; 8 is even.

  3. Rule: cubes of odd numbers, one cube of an even number.

Answer

512

Type 6occasional3 practice Q

Number forms (n²+1, n³−1, n(n+1))

How to spot it:

Numbers sit just beside squares or cubes (50, 65, 82), or are products of consecutive numbers (30, 42, 56).

n2+1,  n3−1,  n(n+1)n^2+1,\; n^3-1,\; n(n+1)
Method
  1. Check each number against the nearest square and cube.

  2. Write it as a square ± 1, a cube ± 1, or n × (n + 1).

  3. The number that does not fit the common form is odd.

Why it works:

The setter builds three numbers from one formula and changes the sign for the fourth.

Try this

Find the odd one: 50, 65, 82, 99.

Show solution
  1. 50 = 7² + 1, 65 = 8² + 1, 82 = 9² + 1.

  2. 99 = 10² − 1, a square MINUS one, not plus one.

  3. Rule: three are n² + 1, one is n² − 1.

Answer

99

09

Formula sheet

Divisibility by 11
∣Sodd−Seven∣≡0(mod11)|S_{odd} - S_{even}| \equiv 0 \pmod{11}

S = sums of alternate digits from the left; e.g. 2728: (2+2) − (7+8) = −11, divisible.

Perfect square endings
n2∈{0,1,4,5,6,9}n^2 \in \{0,1,4,5,6,9\}

A square never ends in 2, 3, 7 or 8 — instant elimination.

Digit sum rule for 9
9∣n  ⟺  S(n)≡0(mod9)9 \mid n \iff S(n) \equiv 0 \pmod 9

S(n) is the sum of the digits of n.

10

Shortcuts that save time

⚡ Last-digit scan for squares

Squares end only in 0, 1, 4, 5, 6 or 9. An option ending in 2, 3, 7 or 8 cannot be a square. For the rest, place the number between two known squares.

Example

Find the odd one: 144, 324, 576, 500.

Show solution
  1. 144 = 12², 324 = 18², 576 = 24². All are squares.

  2. 500 sits between 222=48422^2 = 484 and 232=52923^2 = 529.

  3. So 500 is not a square.

Answer

500

⚡ Prime check by small divisors

To test a number under 200 for prime, divide only by 2, 3, 5, 7, 11 and 13. Memorise the fake-prime list: 51, 57, 87, 91, 119, 133, 143, 161.

Example

Find the odd one: 17, 51, 41, 43.

Show solution
  1. 17, 41 and 43 have no divisor among 2, 3, 5, 7. They are prime.

  2. 51 divides by 3: 51 = 3 × 17.

  3. Rule: three primes, one composite.

Answer

51

11

Mistakes to avoid

Where most students lose marks on this subtopic.

Mistake 01

Treating 1 as prime, or 2 as composite.

1 is neither prime nor composite. 2 is the only even prime.

Mistake 02

Calling 56 odd among multiples of 7 without checking.

56 is also a multiple of 7. Check the rule on every option before marking.

Mistake 03

Stopping at "all are odd" or "all are even".

Parity is almost never the whole rule when all four match. Look at roots or digits next.

Mistake 04

Forgetting a number can be both square and cube (64, 729).

Test both powers before calling a number odd.

Mistake 05

Marking a valid square for the wrong reason in two-level questions.

Write the deeper rule (roots prime, roots odd) and check it on all four.

12

Quick revision

Read this the night before the exam.

  • Test order: prime, square/cube, multiple, digits, parity, forms.

  • Fake primes: 51, 57, 87, 91, 119, 133, 143, 161.

  • Squares to 30230^2, cubes to 12312^3. Squares never end in 2, 3, 7, 8.

  • Digit sum for 3 and 9; alternate-digit difference for 11.

  • Near-forms: n2±1n^2 \pm 1, n3±1n^3 \pm 1, n(n+1)n(n+1).

  • First rule fits all four? Check the roots, then parity.

13

Practice: 20 questions

Sets of 10, mixed across the question types above. Every answer has a step-by-step explanation.

Topic test · 10 questions

Suggested time 5 min · wrong answers go to your mistake notebook automatically.