Question 1

Choose the odd number: Three of the numbers are exact multiples of 17; one is not. Identify the number that is not divisible by 17.

Accepted Answer

Introduction / Context: This classification focuses on a specific divisor (17). The outlier is the one not divisible by 17. Given Data / Assumptions: Candidates: 68, 88, 102, 238. Recall: 17 * 4 = 68; 17 * 6 = 102; 17 * 14 = 238. Concept / Approach: Check division by 17 using simple multiples or quick arithmetic. Step-by-Step Solution: 68 = 17 * 4 → divisible.102 = 17 * 6 → divisible.238 = 17 * 14 → divisible.88 ÷ 17 ≈ 5.176… → not divisible. Verification / Alternative check: Multiply back: 17 * 5 = 85 and 17 * 6 = 102; 88 is between these and hence not a multiple. Why Other Options Are Wrong: They are exact multiples of 17 and follow the majority rule. Common Pitfalls: Relying on rough estimation without confirming adjacent multiples (85 and 102). Final Answer: 88 is not divisible by 17 and is the odd number.

Question 2

Find the odd number: Among 18, 30, 26, and 24, three are multiples of 6 while one is not. Identify the non-multiple of 6.

Accepted Answer

Introduction / Context: The majority are divisible by 6 (i.e., by both 2 and 3). The outlier fails divisibility by 3. Given Data / Assumptions: Options: 18, 30, 26, 24. Multiple of 6 ⇔ divisible by 2 and by 3. Concept / Approach: Apply digit-sum test for 3 and check evenness. Step-by-Step Solution: 18 → even; 1 + 8 = 9 → multiple of 6.30 → even; 3 + 0 = 3 → multiple of 6.24 → even; 2 + 4 = 6 → multiple of 6.26 → even; 2 + 6 = 8 → not divisible by 3 → not a multiple of 6. Verification / Alternative check: Direct division shows only 26/6 is non-integer. Why Other Options Are Wrong: They satisfy divisibility by both 2 and 3. Common Pitfalls: Equating evenness with being a multiple of 6; the 3-test is also required. Final Answer: 26 is the odd number.

Question 3

Choose the odd number: Among 43, 53, 63, and 73, three are prime and one is composite. Identify the composite number.

Accepted Answer

Introduction / Context: Most numbers in this set are primes; the single composite is the odd one out. Given Data / Assumptions: Set: 43, 53, 63, 73. Primes around 40–80 are commonly tested. Concept / Approach: Use digit-sum rule for 3 and quick small-prime checks. Step-by-Step Solution: 63 → 6 + 3 = 9 → divisible by 3 → composite.43 → not divisible by 2, 3, 5 → prime.53 → not divisible by 2, 3, 5 → prime.73 → not divisible by 2, 3, 5 → prime. Verification / Alternative check: Direct division by 7: 63 = 7 * 9 confirms compositeness. Why Other Options Are Wrong: They lack non-trivial divisors and are prime. Common Pitfalls: Assuming “ends with 3” implies divisibility by 3; the rule concerns digit sums, not the last digit. Final Answer: 63 is composite and thus the odd number.

Question 4

Identify the odd number: Exactly one of 26, 124, 728, and 64 is both a perfect square and a perfect cube (i.e., a perfect sixth power). Choose that distinct number.

Accepted Answer

Introduction / Context: Some integers are “perfect powers,” such as perfect squares n^2, perfect cubes n^3, or numbers that are both (n^6). Among these four, one number uniquely satisfies both square and cube properties. Given Data / Assumptions: Candidates: 26, 124, 728, 64. 64 = 8^2 and 4^3 (also 2^6). Concept / Approach: Check if any number matches known squares or cubes. Being both a square and a cube means it is a perfect sixth power. Step-by-Step Solution: 64 → 8^2 and 4^3 → both a square and a cube.26 → not a square (5^2 = 25, 6^2 = 36) and not a cube (2^3 = 8, 3^3 = 27).124 → not a square (11^2 = 121, 12^2 = 144) and not a cube (4^3 = 64, 5^3 = 125).728 → not a square (26^2 = 676, 27^2 = 729) and not a cube (8^3 = 512, 9^3 = 729). Verification / Alternative check: Prime factorization: 64 = 2^6 has exponents multiples of both 2 and 3, confirming square and cube status simultaneously. Why Other Options Are Wrong: They are neither perfect squares nor perfect cubes. Common Pitfalls: Assuming proximity to a well-known square or cube (e.g., 728 vs 729) implies equality. Final Answer: 64 is the number that is both a perfect square and a perfect cube.

Question 5

Find the odd number: In 49, 140, 112, and 97, three are multiples of 7 while one is not. Identify the non-multiple of 7.

Accepted Answer

Introduction / Context: This item distinguishes a non-multiple of 7 among clear multiples of 7. Given Data / Assumptions: Numbers: 49, 140, 112, 97. 49 = 7^2; 140 = 7 * 20; 112 = 7 * 16. Concept / Approach: Confirm multiples of 7 by division or by recognizing standard products. Step-by-Step Solution: 49 → multiple of 7.140 → multiple of 7.112 → multiple of 7.97 → not divisible by 7 (7 * 13 = 91; 7 * 14 = 98). Verification / Alternative check: 97 is prime; immediate neighbors 91 and 98 show it cannot be 7k. Why Other Options Are Wrong: They are exact multiples of 7 and fit the majority property. Common Pitfalls: Mistaking 98 as present; careful reading avoids misselection. Final Answer: 97 is not a multiple of 7 and is the odd number.

Question 6

Choose the odd number: 121, 169, and 289 are squares of primes (11^2, 13^2, 17^2). One number is the square of a composite (15^2). Identify that odd square.

Accepted Answer

Introduction / Context: All four are perfect squares, but three are squares of primes while one is the square of a composite integer, making it the odd one out. Given Data / Assumptions: 121 = 11^2 (prime base). 169 = 13^2 (prime base). 289 = 17^2 (prime base). 225 = 15^2 (composite base; 15 = 3 * 5). Concept / Approach: Classify by the nature of the base whose square forms the number: prime versus composite. Step-by-Step Solution: Identify bases: 11, 13, 15, 17.Check primality of each base: 11, 13, 17 are prime; 15 is composite.Therefore, 225 is distinct as a square of a composite. Verification / Alternative check: Factor 225 = 15 * 15 = (3 * 5) * (3 * 5) = 3^2 * 5^2; the base 15 is not prime. Why Other Options Are Wrong: 121, 169, and 289 are squares of prime bases, matching the majority rule. Common Pitfalls: Assuming all squares are alike without considering the primality of the base. Final Answer: 225 is the odd square (square of a composite).

Question 7

Identify the odd number: Among 119, 154, 85, and 51, three are multiples of 17 while one is not. Choose the non-multiple of 17.

Accepted Answer

Introduction / Context: This set highlights a specific divisor (17). The outlier is the number not divisible by 17. Given Data / Assumptions: 119 = 7 * 17; 85 = 5 * 17; 51 = 3 * 17. 154 is near 153 (9 * 17) and 170 (10 * 17). Concept / Approach: Compare to adjacent multiples of 17 to test divisibility quickly. Step-by-Step Solution: 119 → 17 * 7 → divisible.85 → 17 * 5 → divisible.51 → 17 * 3 → divisible.154 → not equal to 17 * k for integer k (between 153 and 170) → not divisible. Verification / Alternative check: Compute 154 ÷ 17 ≈ 9.058… non-integer. Hence it is the outlier. Why Other Options Are Wrong: Each is an exact multiple of 17. Common Pitfalls: Assuming “nearby” to a multiple implies divisibility; exact equality is required. Final Answer: 154 is not divisible by 17 and is the odd number.

Question 8

Choose the odd number: Three numbers are divisible by 4; one is not. Identify the number that is not a multiple of 4.

Accepted Answer

Introduction / Context: Divisibility by 4 can be checked using the last two digits. The outlier fails this test. Given Data / Assumptions: Set: 64, 208, 316, 118. Rule: A number is divisible by 4 if its last two digits form a number divisible by 4. Concept / Approach: Inspect the last two digits and apply the rule quickly. Step-by-Step Solution: 64 → last two digits 64 → divisible by 4.208 → last two digits 08 → divisible by 4.316 → last two digits 16 → divisible by 4.118 → last two digits 18 → not divisible by 4. Verification / Alternative check: Direct division: 64/4 = 16, 208/4 = 52, 316/4 = 79, while 118/4 = 29.5 (non-integer). Why Other Options Are Wrong: They satisfy the divisibility-by-4 condition. Common Pitfalls: Checking only evenness; divisibility by 4 is stricter than by 2. Final Answer: 118 is not divisible by 4 and is the odd number.

Question 9

Find the odd number: Among 143, 171, 117, and 195, three are multiples of 13 while one is not. Identify the non-multiple of 13.

Accepted Answer

Introduction / Context: This classification tests recognition of multiples of 13. The single non-multiple is the outlier. Given Data / Assumptions: 143 = 11 * 13; 117 = 9 * 13; 195 = 15 * 13. 171 is close to 13 * 13 = 169 but is not equal. Concept / Approach: Compare each candidate with nearby products of 13. Step-by-Step Solution: 143 → multiple of 13.117 → multiple of 13.195 → multiple of 13.171 → not a multiple of 13 (169 and 182 are the adjacent multiples). Verification / Alternative check: 171 ÷ 13 ≈ 13.153… non-integer; hence not divisible by 13. Why Other Options Are Wrong: They are exact multiples of 13. Common Pitfalls: Confusing proximity (171 near 169) with exact divisibility. Final Answer: 171 is not a multiple of 13 and is the odd number.

Question 10

Choose the odd number: In 17, 44, 29, and 13, three are prime while one is composite. Identify the composite number.

Accepted Answer

Introduction / Context: The task is to separate a composite number from a set otherwise consisting of primes. Given Data / Assumptions: Numbers: 17, 44, 29, 13. Primes have exactly two divisors. Concept / Approach: Check quick divisibility: even > 2 implies composite; then confirm that the others lack small prime divisors. Step-by-Step Solution: 44 → even and > 2 → composite (2 * 22 = 4 * 11).17 → prime.29 → prime.13 → prime. Verification / Alternative check: Prime checks up to sqrt(n) for 17, 29, 13 show no non-trivial factors. Why Other Options Are Wrong: They are prime and consistent with the majority property. Common Pitfalls: Assuming a larger value is more likely composite; size alone is not decisive. Final Answer: 44 is composite and thus the odd number.

Question 11

Classification – Odd one out (divisibility by 3) Three of the following integers are divisible by 3, while exactly one fails the divisibility-by-3 rule. Identify the number that is not divisible by 3 and mark it as the odd one out.

Accepted Answer

Introduction / Context: Odd-one-out problems often revolve around a single arithmetic property that three options share while one does not. Here, that shared property is divisibility by 3. The quickest path to a reliable answer is to apply the digit-sum test, a standard technique taught in mental math and elementary number theory. Given Data / Assumptions: Candidates: 63, 126, 215, 342 Target property: divisibility by 3 We assume ordinary base-10 integers with no hidden constraints. Concept / Approach: A number is divisible by 3 if and only if the sum of its digits is divisible by 3. This rule dramatically reduces calculation because we never need to perform long division; we only add digits and check membership in {0, 3, 6, 9, 12, 15, …}. When three items satisfy this rule and one item fails it, the failing item becomes the odd one out. Step-by-Step Solution: 63 → 6 + 3 = 9 → divisible by 3 → fits pattern.126 → 1 + 2 + 6 = 9 → divisible by 3 → fits pattern.342 → 3 + 4 + 2 = 9 → divisible by 3 → fits pattern.215 → 2 + 1 + 5 = 8 → not divisible by 3 → breaks pattern. Verification / Alternative check: Direct division corroborates the test: 63/3 = 21, 126/3 = 42, 342/3 = 114, all integers; 215/3 = 71 remainder 2, not an integer. The two methods agree, so our classification is secure. Why Other Options Are Wrong: 63: Clean multiple of 3; not the odd item. 126: Clean multiple of 3; not the odd item. 342: Clean multiple of 3; not the odd item. None of these: There is a unique non-multiple (215), so this distractor is incorrect. Common Pitfalls: Do not confuse “evenness” with divisibility by 3. Being even (like 342) gives no information about divisibility by 3. Also avoid mental shortcuts such as “numbers ending in 5 are divisible by 5 only”; although true about divisibility by 5, it says nothing about 3—always compute the digit sum. Finally, beware of assuming that the largest or smallest number is special; size is irrelevant here. The only decisive feature is the digit-sum rule, which uniquely flags 215 as the outlier. Final Answer: 215

Question 12

Classification – Odd one out (divisibility by 16) Among the following integers, three are exact multiples of 16. Identify the number that is not divisible by 16 and mark it as the odd one out.

Accepted Answer

Introduction / Context: This classification item hinges on recognizing a higher power-of-two factor: 16 = 2^4. While many candidates quickly see whether a number is even, fewer immediately confirm the presence of four factors of 2. Practicing this check is valuable for fast elimination in test settings. Given Data / Assumptions: Candidates: 272, 210, 240, 304 Target property: divisibility by 16 We work with ordinary base-10 integers. Concept / Approach: There are two practical methods: (1) perform integer division by 16 and check for remainder 0; or (2) factor out powers of 2 and ensure at least four such factors are present. Because 16 is a neat benchmark, mental division is typically quick with these values. Step-by-Step Solution: 272 / 16 = 17 → integer → divisible by 16.240 / 16 = 15 → integer → divisible by 16.304 / 16 = 19 → integer → divisible by 16.210 / 16 = 13.125 → not an integer → not divisible by 16. Verification / Alternative check: Prime-factor view: 272 = 16 * 17, 240 = 16 * 15, 304 = 16 * 19 all contain 2^4; 210 = 2 * 3 * 5 * 7 has only one factor of 2 and therefore cannot reach 2^4. Why Other Options Are Wrong: 272: Multiple of 16 → fits pattern. 240: Multiple of 16 → fits pattern. 304: Multiple of 16 → fits pattern. None of these: A single non-multiple exists (210). Common Pitfalls: Equating “even” with “divisible by 16.” Many even numbers (including 210) lack enough factors of 2. Always confirm using 16 directly or count powers of 2. Final Answer: 210

Question 13

Classification – Odd one out (perfect squares) Three of the following numbers are perfect squares. Identify the number that is not a perfect square and mark it as the odd one out.

Accepted Answer

Introduction / Context: Perfect squares are common anchors in classification puzzles. Rapid recognition of standard squares (12^2 = 144, 13^2 = 169, 18^2 = 324) allows you to isolate a non-square quickly, which is often the intended odd element in such sets. Given Data / Assumptions: Candidates: 144, 169, 288, 324 Target: find the unique non-square. Concept / Approach: Compare each value to nearby known squares and confirm exact equality. If a number sits strictly between two consecutive squares, it cannot be a perfect square. This method avoids unnecessary root calculations. Step-by-Step Solution: 144 = 12^2 → square.169 = 13^2 → square.324 = 18^2 → square.288 lies between 16^2 = 256 and 17^2 = 289, and is not equal to either → not a square. Verification / Alternative check: Square-root estimate: sqrt(288) ≈ 16.97, not an integer. Therefore 288 is not a perfect square, while the others match exact k^2 values. Why Other Options Are Wrong: 144: Exact square. 169: Exact square. 324: Exact square. None of these: The unique non-square is 288. Common Pitfalls: Believing that proximity to a square implies squareness. Only exact equality to k^2 counts. Final Answer: 288

Question 14

Classification – Odd one out (divisibility by 6) Exactly three of the numbers below are divisible by 6 (both even and divisible by 3). Identify the number that is not divisible by 6 and mark it as the odd one out.

Accepted Answer

Introduction / Context: Divisibility by 6 requires two simultaneous conditions: divisibility by 2 (evenness) and by 3 (digit-sum multiple of 3). This dual-check framework is a frequent theme in classification items because it forces candidates to apply two quick rules instead of one. Given Data / Assumptions: Candidates: 120, 72, 108, 98 Goal: find the sole non-multiple of 6. Concept / Approach: First check evenness; then check the digit-sum for divisibility by 3. If both hold, the number is a multiple of 6. If either fails, it is not a multiple of 6. Step-by-Step Solution: 120 → even; 1 + 2 + 0 = 3 → divisible by 3 → 120 is a multiple of 6.72 → even; 7 + 2 = 9 → divisible by 3 → 72 is a multiple of 6.108 → even; 1 + 0 + 8 = 9 → divisible by 3 → 108 is a multiple of 6.98 → even; 9 + 8 = 17 → not divisible by 3 → not a multiple of 6. Verification / Alternative check: Direct division confirms: 120/6 = 20, 72/6 = 12, 108/6 = 18 (all integers). 98/6 = 16 remainder 2, not an integer. Methods agree. Why Other Options Are Wrong: 120: Satisfies both conditions; not odd one out. 72: Satisfies both conditions; not odd one out. 108: Satisfies both conditions; not odd one out. None of these: There is a clear unique non-multiple (98). Common Pitfalls: Stopping after the evenness check. Remember: 6 requires both 2 and 3 as factors. Final Answer: 98

Question 15

Classification – Odd one out (multiples of 7) Three of the following integers are clean multiples of 7. Identify the number that is not a multiple of 7 and mark it as the odd one out.

Accepted Answer

Introduction / Context: Multiples of a small prime such as 7 are common in classification exercises. Faster solutions come from either recalling times-table anchors or performing small divisions without long arithmetic. Given Data / Assumptions: Candidates: 28, 35, 42, 65 Property in focus: divisibility by 7 Concept / Approach: Check each candidate by quick division or factor recall. 28 = 7 * 4, 35 = 7 * 5, 42 = 7 * 6 are well-known. The remaining value should fail the 7-multiple test to become the odd one out. Step-by-Step Solution: 28 → 7 * 4 → multiple of 7.35 → 7 * 5 → multiple of 7.42 → 7 * 6 → multiple of 7.65 ÷ 7 = 9 remainder 2 → not a multiple of 7. Verification / Alternative check: Prime-factor view: 65 = 5 * 13 contains no factor 7, confirming it is the outlier. Why Other Options Are Wrong: 28: Multiple of 7. 35: Multiple of 7. 42: Multiple of 7. None of these: A single non-multiple exists (65). Common Pitfalls: Assuming “ends with 5” automatically signals special status. Ending with 5 relates to divisibility by 5, not by 7; always test the actual target factor. Final Answer: 65

Question 16

Classification – Odd one out (unique multiple of 13) Among the following numbers that all end with the digit 5, exactly one is a multiple of 13. Identify that number and mark it as the odd one out.

Accepted Answer

Introduction / Context: When several options share a surface feature (here, all end with 5 → all are multiples of 5), the true classification hinge is often a secondary factor. In this item, that hidden factor is divisibility by 13. Only one number in the set carries 13 as a factor, which makes it the odd one out even though all options are multiples of 5. Given Data / Assumptions: Candidates: 115, 145, 195, 75 Shared surface property: each ends with 5 (therefore divisible by 5) Hidden discriminant: divisibility by 13 Concept / Approach: First, acknowledge the commonality (×5). Next, test each value for 13. Efficient checks: 13 * 15 = 195; 13 * 11 = 143 (near 145 but not equal); 13 * 9 = 117 (near 115 but not equal); 13 does not divide 75. Thus 195 uniquely contains 13 as a factor. Step-by-Step Solution: 115 = 5 * 23 → not divisible by 13.145 = 5 * 29 → not divisible by 13 (13 * 11 = 143, 13 * 12 = 156).75 = 3 * 5 * 5 → not divisible by 13.195 = 13 * 15 → divisible by 13 → odd one out. Verification / Alternative check: Modular test: for k ending in 5, k ≡ 0 (mod 5). Compute k (mod 13) to isolate the one ≡ 0 (mod 13): 195 (mod 13) = 0; the others produce nonzero residues. Why Other Options Are Wrong: 115: Not divisible by 13. 145: Not divisible by 13. 75: Not divisible by 13. None of these: A single 13-multiple exists (195). Common Pitfalls: Stopping after noticing that all end with 5, which would make the set appear homogeneous. Always look for a deeper factor that can create a unique outlier. Final Answer: 195

Question 17

Classification – Odd one out (divisibility by 6) Three of the given numbers are multiples of 6. Identify the number that is not divisible by 6 and mark it as the odd one out.

Accepted Answer

Introduction / Context: A multiple of 6 must be divisible by both 2 and 3. This two-part criterion quickly separates qualifying values from look-alikes. In sets where three satisfy the rule, the lone non-qualifier becomes the odd element. Given Data / Assumptions: Candidates: 6, 24, 64, 120 We test the 2-and-3 divisibility condition for 6. Concept / Approach: Check evenness (divisible by 2) and the digit-sum test for 3. A number must pass both to be a multiple of 6. Step-by-Step Solution: 6 → even and 6/3 = 2 → multiple of 6.24 → even; 2 + 4 = 6 → divisible by 3 → multiple of 6.120 → even; 1 + 2 + 0 = 3 → divisible by 3 → multiple of 6.64 → even; 6 + 4 = 10 → not divisible by 3 → not a multiple of 6. Verification / Alternative check: Direct division: 6/6 = 1, 24/6 = 4, 120/6 = 20 (integers). 64/6 = 10 remainder 4 → not an integer. Why Other Options Are Wrong: 6: Multiple of 6. 24: Multiple of 6. 120: Multiple of 6. None of these: There is a unique non-multiple (64). Common Pitfalls: Assuming that any even number qualifies. Evenness alone is insufficient; the 3-divisibility condition must also hold. Final Answer: 64

Question 18

Classification – Odd one out (prime vs composite) Among the following integers, three are prime while exactly one is composite. Identify the composite number and mark it as the odd one out.

Accepted Answer

Introduction / Context: Prime recognition is a core classification skill. Test-makers frequently present three unmistakable primes along with a composite that yields to a small-prime divisibility check. Spotting the composite quickly is the key to speed. Given Data / Assumptions: Candidates: 53, 67, 73, 87 We separate primes from composites using small divisibility tests. Concept / Approach: Check divisibility by 2, 3, 5 first, then 7 and 11 as needed. The digit-sum rule is especially useful for screening multiples of 3. Step-by-Step Solution: 87 → 8 + 7 = 15 → divisible by 3 → composite (3 * 29).53 → not divisible by 2, 3, 5, 7 → prime.67 → not divisible by 2, 3, 5, 7 → prime.73 → not divisible by 2, 3, 5, 7 → prime. Verification / Alternative check: Trial division up to sqrt(87) < 10 confirms 87 has factor 3. The others have no small prime divisors within their square-root bounds. Why Other Options Are Wrong: 53: Prime, so not the odd composite. 67: Prime, so not the odd composite. 73: Prime, so not the odd composite. None of these: There is a single composite (87). Common Pitfalls: Confusing “odd” with “prime.” Many odd numbers are composite; always apply check rules. Final Answer: 87

Question 19

Classification – Odd one out (parity) Exactly one of the following integers is even/odd-inconsistent with the others. Identify the outlier based on parity and mark it as the odd one out.

Accepted Answer

Introduction / Context: Parity (even vs odd) is the most elementary but still legitimate basis for many classification questions. Because parity is glance-detectable from the last digit, it offers a near-instant path to the odd element when three values share one parity and one value has the opposite. Given Data / Assumptions: Set: 426, 369, 279, 159 Property examined: parity (even/odd) Concept / Approach: A number is even if its last digit is 0, 2, 4, 6, or 8; otherwise it is odd. Scan the terminal digits and tally the parities to find the minority class (the odd one out). Step-by-Step Solution: 426 → ends with 6 → even.369 → ends with 9 → odd.279 → ends with 9 → odd.159 → ends with 9 → odd. Verification / Alternative check: Basic divisibility by 2: only 426 is divisible by 2. Therefore, 426 is the sole even number among three odds. Why Other Options Are Wrong: 369: Odd like two others; not unique. 279: Odd like two others; not unique. 159: Odd like two others; not unique. None of these: There is a clear even outlier (426). Common Pitfalls: Overcomplicating with factors of 3 or 9. While those patterns exist, parity alone yields a unique classification here and is the intended quick route. Final Answer: 426

Question 20

Classification – Odd one out (multiples of 13) In the following set, three numbers are multiples of 13 while one is not. Identify the non-multiple of 13 and mark it as the odd one out.

Accepted Answer

Introduction / Context: Recognizing multiples of medium primes (like 13) is a useful classification skill. Many exam sets include two or three crisp multiples and a single non-multiple to create a unique outlier. Given Data / Assumptions: Candidates: 91, 143, 247, 215 Property checked: divisibility by 13 Concept / Approach: Recall familiar multiples: 13 * 7 = 91, 13 * 11 = 143, 13 * 19 = 247. If three values match this pattern, the leftover non-multiple is the odd element. Step-by-Step Solution: 91 = 13 * 7 → multiple of 13.143 = 13 * 11 → multiple of 13.247 = 13 * 19 → multiple of 13.215 = 5 * 43 → not a multiple of 13. Verification / Alternative check: Divide directly by 13: 91/13 = 7, 143/13 = 11, 247/13 = 19 (integers). 215/13 = 16 remainder 7, so not an integer. Therefore, 215 is the unique non-multiple. Why Other Options Are Wrong: 91: Multiple of 13. 143: Multiple of 13. 247: Multiple of 13. None of these: There is a single non-multiple (215). Common Pitfalls: Confusing “ends with 5” as an indicator of special status. Ending in 5 reveals divisibility by 5, not by 13. Always test the target factor explicitly. Final Answer: 215

Classification Questions