Question 1

Classification – Odd one out (perfect square among non-squares) Among the following integers, three are not perfect squares while exactly one is a perfect square. Identify the perfect square and mark it as the odd one out.

Accepted Answer

Introduction / Context:Odd-one-out items often target quick recognition of perfect squares. Being able to spot classic squares like 29^2 = 841 lets you classify efficiently without lengthy computation.Given Data / Assumptions:Candidates: 626, 841, 962, 1090Goal: find the single perfect squareAll values are base-10 integers with no extra conditionsConcept / Approach:Recall common squares around 30^2 = 900. Specifically, 29^2 = 841 and 31^2 = 961. Compare each option to see whether it equals k^2 for some integer k. If a number sits strictly between two consecutive squares, it is not a perfect square.Step-by-Step Solution:841 = 29^2 → perfect square.626 lies between 25^2 = 625 and 26^2 = 676 → not a square.962 lies between 31^2 = 961 and 32^2 = 1024 → not a square.1090 lies between 33^2 = 1089 and 34^2 = 1156 → not a square.Verification / Alternative check:Square-root estimates: sqrt(841) = 29 exactly; sqrt(626) ≈ 25.0+, sqrt(962) ≈ 31.0+, sqrt(1090) ≈ 33.0+, none integral. Only 841 matches an integer root, confirming it as the unique square.Why Other Options Are Wrong:626: Not equal to any k^2.962: Not equal to any k^2.1090: Not equal to any k^2.None of these: There is one clear square (841).Common Pitfalls:Assuming that being close to a square (e.g., 626 near 625) implies squareness. Only exact equality to k^2 qualifies.Final Answer:841

Question 2

Classification – Odd one out (unique prime among composites) Exactly one of the following integers is prime while the others are composite. Identify the prime number and mark it as the odd one out.

Accepted Answer

Introduction / Context:Prime-versus-composite classification is common. The fastest route is to eliminate obvious composites by small-prime checks (2, 3, 5, 7, 11, 13, …) and then test the surviving candidate up to its square root.Given Data / Assumptions:Candidates: 488, 776, 667, 929We look for the single prime.Concept / Approach:First discard even numbers (composite if > 2): 488 and 776 are even. Next check 667: it equals 23 * 29 (composite). The remaining number 929 must be tested up to sqrt(929) ≈ 30.5 against primes 3, 5, 7, 11, 13, 17, 19, 23, 29. None divides 929 evenly, making it prime.Step-by-Step Solution:488 → even → composite.776 → even → composite.667 → 23 * 29 → composite.929 → not divisible by 3, 5, 7, 11, 13, 17, 19, 23, 29 → prime.Verification / Alternative check:Trial division up to 29 confirms no exact divisor for 929; therefore it is prime. All others have explicit composite factorizations or are even > 2.Why Other Options Are Wrong:488: Composite (even).776: Composite (even).667: Composite (23 * 29).None of these: There is a unique prime (929).Common Pitfalls:Stopping at parity checks only. While parity helps, you must also test odd candidates against small primes to avoid mislabeling.Final Answer:929

Question 3

Classification – Odd one out (divisibility by 3) Three of these four-digit numbers are not divisible by 3, while exactly one is divisible by 3. Identify the multiple of 3 and mark it as the odd one out.

Accepted Answer

Introduction / Context:Digit-sum divisibility is a staple in reasoning tests. For 3, a number is divisible if its digit sum is a multiple of 3. Here, exactly one value meets this rule, making it the odd one out.Given Data / Assumptions:Candidates: 3216, 2338, 3205, 2015Property checked: divisibility by 3 using digit-sumConcept / Approach:Compute digit sums and check membership in {0, 3, 6, 9, 12, …}. Only the number with a digit sum divisible by 3 is a multiple of 3.Step-by-Step Solution:3216 → 3 + 2 + 1 + 6 = 12 → divisible by 3.2338 → 2 + 3 + 3 + 8 = 16 → not divisible by 3.3205 → 3 + 2 + 0 + 5 = 10 → not divisible by 3.2015 → 2 + 0 + 1 + 5 = 8 → not divisible by 3.Verification / Alternative check:Direct division: 3216/3 = 1072 (integer). The others leave remainders upon division by 3.Why Other Options Are Wrong:2338: Fails digit-sum test.3205: Fails digit-sum test.2015: Fails digit-sum test.None of these: There is one clear multiple of 3 (3216).Common Pitfalls:Judging by evenness or size instead of using the correct rule. Only the digit-sum test matters for 3.Final Answer:3216

Question 4

Classification – Odd one out (multiples of 11) Three of the following integers are exact multiples of 11. Identify the number that is not a multiple of 11 and mark it as the odd one out.

Accepted Answer

Introduction / Context:Recognizing multiples of 11 can be done by memory for common values or by using quick division. Many test items include three clean multiples and a single non-multiple to create a unique outlier.Given Data / Assumptions:Candidates: 572, 671, 264, 427Property checked: divisibility by 11Concept / Approach:Use direct division or recall: 11 * 24 = 264, 11 * 52 = 572, 11 * 61 = 671. If three values match, the leftover is the non-multiple.Step-by-Step Solution:264 = 11 * 24 → multiple.572 = 11 * 52 → multiple.671 = 11 * 61 → multiple.427 / 11 = 38 remainder 9 → not a multiple.Verification / Alternative check:Applying the alternating-sum rule for 11 also works, but direct factor matches already confirm the classification.Why Other Options Are Wrong:264: Clean multiple of 11.572: Clean multiple of 11.671: Clean multiple of 11.None of these: There is exactly one non-multiple (427).Common Pitfalls:Overlooking that 671 is 11 * 61 due to unfamiliarity with larger 11-tables. Memorizing a few key anchors helps.Final Answer:427

Question 5

Classification – Odd one out (three perfect squares vs one non-square) Among the following numbers, three are perfect squares and one is not. Identify the non-square and mark it as the odd one out.

Accepted Answer

Introduction / Context:This item checks rapid recognition of standard squares: 12^2 = 144, 13^2 = 169, 14^2 = 196. Spotting the outsider reinforces square benchmarks crucial for speed in classification tasks.Given Data / Assumptions:Candidates: 144, 169, 196, 210Goal: find the single non-square.Concept / Approach:Confirm equality to k^2 for small integers. If a value lies strictly between consecutive squares, it cannot be a perfect square.Step-by-Step Solution:144 = 12^2 → square.169 = 13^2 → square.196 = 14^2 → square.210 is between 14^2 = 196 and 15^2 = 225 → not a perfect square.Verification / Alternative check:Square-root estimation: sqrt(210) ≈ 14.49, not an integer, confirming non-square status.Why Other Options Are Wrong:144: Exact square.169: Exact square.196: Exact square.None of these: 210 is uniquely non-square.Common Pitfalls:Mistaking “close to a square” as sufficient. Only exact equality qualifies.Final Answer:210

Question 6

Classification – Odd one out (three perfect squares vs one non-square) In the set below, three numbers are perfect squares, whereas one is not a perfect square. Identify the non-square as the odd one out.

Accepted Answer

Introduction / Context:A classic classification pattern contrasts perfect squares with a lone non-square. Recognizing 4^2 = 16, 5^2 = 25, and 7^2 = 49 quickly reveals the outlier.Given Data / Assumptions:Set: 16, 25, 35, 49We must find the single non-square.Concept / Approach:Match candidates to familiar squares. If a value does not equal k^2 for any integer k, it is not a perfect square.Step-by-Step Solution:16 = 4^2 → square.25 = 5^2 → square.49 = 7^2 → square.35 is between 5^2 = 25 and 6^2 = 36 → not a square.Verification / Alternative check:sqrt(35) ≈ 5.916… is non-integral, confirming 35 is not a perfect square.Why Other Options Are Wrong:16: Exact square.25: Exact square.49: Exact square.None of these: There is clearly one non-square (35).Common Pitfalls:Assuming “odd = non-square.” Many odd numbers are squares (e.g., 25, 49). Always verify with known square facts.Final Answer:35

Question 7

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

Accepted Answer

Introduction / Context:Small-prime multiples (like 7) are frequent in classification items. A quick recall of 7-tables helps isolate the non-multiple instantly.Given Data / Assumptions:Set: 21, 55, 63, 49Property target: divisibility by 7Concept / Approach:Confirm each value against 7: 21 = 7 * 3, 63 = 7 * 9, 49 = 7 * 7. The remaining value fails the test and becomes the outlier.Step-by-Step Solution:21 → 7 * 3 → multiple.63 → 7 * 9 → multiple.49 → 7 * 7 → multiple.55 → 5 * 11 → not a multiple of 7.Verification / Alternative check:Direct division 55/7 gives remainder 6, confirming non-divisibility.Why Other Options Are Wrong:21: Clean multiple of 7.63: Clean multiple of 7.49: Clean multiple of 7.None of these: Exactly one non-multiple exists (55).Common Pitfalls:Confusing “ends with 1 or 3 or 9” as a rule. Ending digit patterns do not determine 7-divisibility; use factors or division.Final Answer:55

Question 8

Classification – Odd one out (parity: only odd vs evens) Among the following five-digit and four-digit integers, three are even while exactly one is odd. Identify the odd number (by parity) and mark it as the odd one out.

Accepted Answer

Introduction / Context:Parity (even/odd) is a rapid classification key. When three values share evenness and one is odd, the parity test immediately yields the outlier with zero arithmetic beyond checking the last digit.Given Data / Assumptions:Set: 72572, 35453, 78378, 46246We test parity only.Concept / Approach:An integer is even if its last digit is 0, 2, 4, 6, or 8; otherwise it is odd. Scan the last digit of each number and select the single odd element.Step-by-Step Solution:72572 → ends with 2 → even.78378 → ends with 8 → even.46246 → ends with 6 → even.35453 → ends with 3 → odd → outlier.Verification / Alternative check:Any further divisibility checks are unnecessary; parity alone uniquely identifies 35453 as different from the other three.Why Other Options Are Wrong:72572: Even; part of the majority class.78378: Even; part of the majority class.46246: Even; part of the majority class.None of these: There is exactly one odd number (35453).Common Pitfalls:Overcomplicating with prime or 11-divisibility rules when a simpler parity rule suffices and yields a unique choice.Final Answer:35453

Question 9

Classification – Odd one out (fixed difference within ordered pairs) Each option is an ordered pair of integers (x, y). In three pairs, the difference y − x equals 11; in exactly one pair, the difference is not 11. Identify the pair that breaks the …

Accepted Answer

Introduction / Context:Classification questions with ordered pairs often hide a consistent arithmetic relation. Here, the governing relation is a fixed difference between the two numbers in each pair. Detecting and verifying that relation reveals the single exception.Given Data / Assumptions:Pairs: (46, 57), (38, 49), (41, 52), (64, 73)We test the difference y − x for each pair.Concept / Approach:Compute y − x for every pair and look for consistency. If most pairs share the same difference and one does not, that pair is the odd element.Step-by-Step Solution:(46, 57): 57 − 46 = 11 → fits pattern.(38, 49): 49 − 38 = 11 → fits pattern.(41, 52): 52 − 41 = 11 → fits pattern.(64, 73): 73 − 64 = 9 → breaks pattern.Verification / Alternative check:Since three independent differences equal 11, the relation is unambiguous; any other arithmetic property would be secondary and unnecessary.Why Other Options Are Wrong:46, 57: Matches difference 11.38, 49: Matches difference 11.41, 52: Matches difference 11.None of these: One clear exception exists (64, 73).Common Pitfalls:Overfitting patterns such as sums or products when a simpler, fixed-difference rule already yields a unique odd one out.Final Answer:64, 73

Question 10

Classification – Odd one out (interval span) Each option denotes a closed interval [a, b]. In three options, the span b − a equals 2; in exactly one option, the span is different. Identify the interval with a different span.

Accepted Answer

Introduction / Context:Intervals can be compared by their span (b − a). Test-makers often keep three spans equal and alter one to create a unique outlier. Identifying the inconsistent span is a quick arithmetic check.Given Data / Assumptions:Intervals: 11–13, 12–14, 17–19, 18–22Interpretation: closed intervals [a, b] with integer endpointsConcept / Approach:Compute b − a for each interval. If three intervals have span 2 and one has a different span, that different one is the odd element. (We compare spans only; whether endpoints are prime is irrelevant here.)Step-by-Step Solution:11–13 → span = 13 − 11 = 2 → fits pattern.12–14 → span = 14 − 12 = 2 → fits pattern.17–19 → span = 19 − 17 = 2 → fits pattern.18–22 → span = 22 − 18 = 4 → breaks pattern.Verification / Alternative check:Even if you looked for prime twins, both 11–13 and 17–19 are twin-prime intervals; however, 12–14 includes composites. The span test alone consistently isolates one unique outlier (18–22), which is the intended basis.Why Other Options Are Wrong:11–13: Span 2.12–14: Span 2.17–19: Span 2.None of these: There is a single different span (18–22).Common Pitfalls:Mixing multiple criteria (e.g., primes and parity) and missing the simplest invariant—span equality.Final Answer:18-22

Question 11

Odd One Out — In the number pairs 20–36, 30–36, 50–56, and 60–66, pick the pair that does not follow the same constant-difference pattern. Explain the rule you used.

Accepted Answer

Introduction / Context:Classification questions on number pairs often rely on a simple arithmetic relation like a constant difference or ratio. The task is to find the one pair that breaks the common rule.Given Data / Assumptions:Pairs provided: 20–36, 30–36, 50–56, 60–66.We interpret “a–b” as an ordered pair (a, b) with a simple relation connecting a to b.We will test the most economical relation first: constant difference (b − a).Concept / Approach:Compute the difference b − a for each pair and compare the results. When three share the same difference and one does not, the odd one is isolated.Step-by-Step Solution:For 20–36: difference = 36 − 20 = 16.For 30–36: difference = 6.For 50–56: difference = 6.For 60–66: difference = 6.Observation: three pairs share the same difference (+6). One pair has a larger, different difference (+16).Verification / Alternative check:Check ratios quickly: 36/20 = 1.8; 36/30 = 1.2; 56/50 = 1.12; 66/60 = 1.1. Ratios are not equal, so the constant-difference pattern is the most coherent commonality among three pairs.Why Other Options Are Wrong:30–36: fits b − a = +6.50–56: fits b − a = +6.60–66: fits b − a = +6.Common Pitfalls:A frequent mistake is to jump to multiplicative patterns when an additive pattern already cleanly explains three of the four options. Always test simplest invariants (difference, parity) first.Final Answer:20-36

Question 12

Odd One Out — In 117–39, 164–41, 198–66, 213–71, three have the same quotient (first ÷ second). Identify the exception and justify.

Accepted Answer

Introduction / Context:Number-pair classification commonly uses division to reveal a hidden constant quotient. Your goal is to find the one pair that breaks the uniform quotient rule.Given Data / Assumptions:Pairs: 117–39, 164–41, 198–66, 213–71.Interpret each as (a, b) and compute a / b.Concept / Approach:If three quotients are equal and one is not, the latter is the odd one out.Step-by-Step Solution:117 / 39 = 3.198 / 66 = 3.213 / 71 = 3.164 / 41 = 4.Thus, three pairs share quotient 3; one pair yields quotient 4.Verification / Alternative check:Confirm by multiplication: 39*3 = 117; 66*3 = 198; 71*3 = 213; 41*3 = 123 ≠ 164 (indeed 41*4 = 164). This reconfirms the outlier.Why Other Options Are Wrong:117–39: follows quotient 3.198–66: follows quotient 3.213–71: follows quotient 3.Common Pitfalls:Checking differences first does not isolate a single outlier (differences vary throughout). Division reveals the clean triad.Final Answer:164-41

Question 13

Odd One Out — Among 25631, 33442, 34424, and 52163, three have the same digit sum. Find the exception and explain.

Accepted Answer

Introduction / Context:Digit-based invariants (sum, product, parity of sum) are classic levers in classification items. We will use digit sums to isolate the odd number.Given Data / Assumptions:Numbers: 25631, 33442, 34424, 52163.Digit sum means adding all decimal digits of the number.Concept / Approach:Compute each digit sum and compare. When three match and one differs, that one is the outlier.Step-by-Step Solution:25631: 2 + 5 + 6 + 3 + 1 = 17.33442: 3 + 3 + 4 + 4 + 2 = 16.34424: 3 + 4 + 4 + 2 + 4 = 17.52163: 5 + 2 + 1 + 6 + 3 = 17.Observation: three values share digit sum 17; one has digit sum 16.Verification / Alternative check:You can also check the sums modulo 3: 17 ≡ 2 (mod 3); 16 ≡ 1 (mod 3). The single residue mismatch also isolates the same item.Why Other Options Are Wrong:25631: digit sum 17; matches the majority.34424: digit sum 17; matches the majority.52163: digit sum 17; matches the majority.Common Pitfalls:Over-searching for complex prime or square properties; the neat invariant here is the digit sum.Final Answer:33442

Question 14

Odd One Out — From the pairs 23–29, 19–25, 13–17, 3–5, select the pair that does not consist of two primes.

Accepted Answer

Introduction / Context:Prime recognition is a staple in classification problems. Here, three pairs contain only primes; the remaining pair includes a composite number.Given Data / Assumptions:Pote ntial primes: 3, 5, 13, 17, 19, 23, 29.25 is 5 * 5 and therefore composite.Concept / Approach:Check each element for primality using small divisibility tests (2, 3, 5, 7) appropriate to the magnitude.Step-by-Step Solution:23–29: both prime.13–17: both prime.3–5: both prime.19–25: 25 is composite (5^2), so this pair breaks the pattern.Verification / Alternative check:Square bounds: 25 is a perfect square; perfect squares greater than 1 are never prime. This confirms the outlier quickly.Why Other Options Are Wrong:23–29: prime pair.13–17: prime pair.3–5: prime pair.Common Pitfalls:Assuming any odd number is prime; 25 is an instructive counterexample.Final Answer:19-25

Question 15

Odd One Out — Consider the pairs (18, 45), (23, 14), (29, 82), (36, 27). Choose the pair that uniquely contains a perfect square and justify.

Accepted Answer

Introduction / Context:When several numerical relations seem plausible (gcd, parity mix, sum), a crisp discriminator is the presence of a perfect square. This item asks you to spot the only pair containing one.Given Data / Assumptions:Pairs: (18, 45), (23, 14), (29, 82), (36, 27).Perfect square means n^2 for some integer n.Concept / Approach:Scan each pair for 4, 9, 16, 25, 36, 49, 64, 81, … The first few squares quickly isolate candidates.Step-by-Step Solution:18: not a square; 45: not a square.23: not; 14: not.29: not; 82: not.36: equals 6^2 (a perfect square); 27: not a square.Therefore, only (36, 27) contains a perfect square.Verification / Alternative check:You might try gcd or “both multiples of 9” to compare pairs. Those checks split the set into two pairs each and do not yield a single outlier. The square test produces a unique selection.Why Other Options Are Wrong:(18, 45): no perfect squares present.(23, 14): no perfect squares present.(29, 82): no perfect squares present.Common Pitfalls:Overfitting with multi-criterion patterns. Favor single, decisive invariants that yield exactly one exception.Final Answer:36, 27

Question 16

Odd One Out — For the pairs 5–2, 19–16, 27–23, 31–28, three have difference 3. Identify the exception and explain.

Accepted Answer

Introduction / Context:Checking constant differences is a fast way to classify number pairs. Here, three pairs share the same difference; one breaks it.Given Data / Assumptions:Pairs: 5–2, 19–16, 27–23, 31–28.We compute difference as a − b or b − a consistently; since each pair shows a larger first number, we use a − b.Concept / Approach:Calculate the difference for each pair and compare.Step-by-Step Solution:5 − 2 = 3.19 − 16 = 3.31 − 28 = 3.27 − 23 = 4.Thus, three differences are 3; one is 4.Verification / Alternative check:Order does not matter for absolute difference: |5 − 2|, |19 − 16|, |31 − 28| are 3; |27 − 23| is 4. The outlier persists.Why Other Options Are Wrong:5–2: fits difference 3.19–16: fits difference 3.31–28: fits difference 3.Common Pitfalls:Switching to ratio checks complicates matters; the simplest invariant suffices here.Final Answer:27-23

Question 17

Odd One Out — In 1:4, 10:24, 8:18, 22:46, three follow the rule “second = 2*first + 2”. Find the exception.

Accepted Answer

Introduction / Context:Linear mappings of the form y = m*x + c are common in pair-classification problems. We identify the lone pair that fails the dominant mapping.Given Data / Assumptions:Pairs: 1:4, 10:24, 8:18, 22:46 (interpreted as (x, y)).Candidate rule observed from three items: y = 2*x + 2.Concept / Approach:Test each pair against y = 2*x + 2 and isolate the pair that violates it.Step-by-Step Solution:For 1:4 — 2*1 + 2 = 4 ✓For 8:18 — 2*8 + 2 = 18 ✓For 22:46 — 2*22 + 2 = 46 ✓For 10:24 — 2*10 + 2 = 22 ≠ 24 ✗Hence, 10:24 is the exception.Verification / Alternative check:Attempting y = 2*x + c with a single c for all pairs yields c = 2 for three items; no single c satisfies 10:24 simultaneously.Why Other Options Are Wrong:1:4 — satisfies y = 2*x + 2.8:18 — satisfies y = 2*x + 2.22:46 — satisfies y = 2*x + 2.Common Pitfalls:Assuming proportion (y = k*x) and missing the additive constant; always test both multiplicative and affine forms.Final Answer:10 : 24

Question 18

Odd One Out — Among 5:9, 7:11, 13:17, 29:31, select the pair that is not a pair of primes.

Accepted Answer

Introduction / Context:Prime versus composite recognition is central to many number-classification items. Three options are prime-prime pairs; one includes a composite number.Given Data / Assumptions:Primes among the listed numbers: 5, 7, 11, 13, 17, 29, 31.9 is composite (3 * 3).Concept / Approach:Check each member of the pair for primality; the pair containing a composite is the outlier.Step-by-Step Solution:5:9 — 5 is prime, 9 is composite → mismatched pair.7:11 — both prime.13:17 — both prime.29:31 — both prime.Verification / Alternative check:All-prime pairs persist even if you test differences (4, 4, 2). The single composite presence is the decisive discriminator.Why Other Options Are Wrong:7:11 — prime pair.13:17 — prime pair.29:31 — prime pair.Common Pitfalls:Confusing “odd” with “prime.” Not all odd numbers are prime; 9 is a counterexample.Final Answer:5 : 9

Question 19

Odd One Out — For 147:741, 253:352, 518:816, 303:303, choose the pair that is not a correct reverse (or self-reverse in case of a palindrome).

Accepted Answer

Introduction / Context:Digit reversal is a frequent theme: “abc” should map to “cba”, while palindromes map to themselves. We must identify the lone mismatch.Given Data / Assumptions:Pairs shown as (left, right).Reversal expectation: reverse(left) = right; for palindromes, left = right.Concept / Approach:Reverse each left-hand number and compare with the right-hand number.Step-by-Step Solution:147 → reverse is 741; matches.253 → reverse is 352; matches.518 → reverse is 815; given right side is 816; does not match.303 → palindrome; reverse is 303; matches.Verification / Alternative check:Confirm each digit ordering explicitly to avoid off-by-one slips (815 vs 816).Why Other Options Are Wrong:147:741 — correct reversal.253:352 — correct reversal.303:303 — correct palindrome mapping.Common Pitfalls:Overlooking palindromic self-mapping or misreading “815” as “816.” Careful digit tracing prevents this.Final Answer:518 : 816

Question 20

Odd One Out — In 22:44, 45:1625, 18:164, 24:464, three right-hand numbers are concatenations of squared digits of the left number. Find the exception.

Accepted Answer

Introduction / Context:A popular mapping for two-digit numbers ab → (a^2)(b^2) forms the right-hand number by concatenating the squares of the digits. We look for the one pair that violates this rule.Given Data / Assumptions:Digits are a (tens) and b (ones) of the left number.Compute a^2 and b^2, then write them next to each other (concatenate) to form the right side.Concept / Approach:Apply the ab → (a^2)(b^2) rule to each left number and compare with the given right number.Step-by-Step Solution:22 → a=2, b=2 → a^2=4, b^2=4 → 44 ✓45 → a=4, b=5 → a^2=16, b^2=25 → 1625 ✓18 → a=1, b=8 → a^2=1, b^2=64 → 164 ✓24 → a=2, b=4 → a^2=4, b^2=16 → expected 416, but given 464 ✗Verification / Alternative check:No reasonable variant (swap or sum) consistently fits the other three pairs, confirming the concatenated-squares rule as dominant and isolating 24:464 as the exception.Why Other Options Are Wrong:22:44 — matches concatenated squares.45:1625 — matches concatenated squares.18:164 — matches concatenated squares.Common Pitfalls:Reading 416 as 461 or 464; precise concatenation order is (a^2)(b^2), not reversed.Final Answer:24 : 464

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