Exponents

24^2 = 576 and 25^2 = 625, so 24^2 + 25^2 = 576 + 625 = 1201. A quick method: compute 24^2 as (25−1)^2 = 25^2 − 2·25 + 1 = 576, then add. The closest distractor 1200 is off by 1 and therefore incorrect. Exam tip: use the identity (n−1)^2 + n^2 = 2n^2 − 2n + 1 for consecutive integers to save time.

Find the two squares first: \(25^2=625\) and \(26^2=676\). Therefore, \(25^2+26^2=625+676=1301\), so option B is correct. Exam tip: calculate each square separately before adding to avoid arithmetic errors.

Given that \(f^3=6859\). Since \(19^3=19\times19\times19=361\times19=6859\), it follows that \(f=19\). Exam tip: For cube equations, check the number whose product taken three times equals the given value.

Direct calculation: \(26^2=676\) and \(27^2=729\). Adding gives \(676+729=1405\). A quick identity: \(a^2+b^2=(a+b)^2-2ab\) — here \((26+27)^2-2\times26\times27=53^2-1404=2809-1404=1405\). Option 1395 typically arises from an arithmetic slip, so it is close but incorrect. Exam tip: use the identity or compute individual squares carefully to avoid addition mistakes under time pressure.

Calculate the two squares first: 27² = 729 and 28² = 784. Therefore, 27² + 28² = 729 + 784 = 1513, so option B is correct. As an exam tip, you can verify 28² using (30 − 2)² = 900 − 120 + 4 = 784 to avoid calculation errors.

Use the identity \(a^2+b^2=(a+b)^2-2ab\) for a quick calculation. With \(a=28, b=29\): \(a+b=57\) and \(ab=28\times29=812\). So \(28^2+29^2=57^2-2\times812=3249-1624=1625\). The close option 1615 is incorrect — it could arise from miscalculating \(29^2\) as 831 instead of 841. Exam tip: either use the identity above or compute \(28^2=784\) and \(29^2=841\) separately and add them to avoid small arithmetic mistakes.

Evaluate the powers first: \(3^7=2187\) and \(3^4=81\). Therefore, \(3^7-3^4=2187-81=2106\), so option A is correct. Option C is only the value of \(3^7\); it does not subtract \(3^4\). Exam tip: when subtracting powers with the same base, subtract their values rather than their exponents.

First calculate the two squares: 29^2 = 841 and 30^2 = 900. Therefore, 29^2 + 30^2 = 841 + 900 = 1741, so option B is correct. In an exam, calculating each square separately before adding helps avoid arithmetic errors.

For division of powers with the same base, use \(a^m \div a^n = a^{m-n}\). Thus, \(2^9 \div 2^4 = 2^{9-4} = 2^5 = 32\). Option B is only the value of \(2^4\); it does not apply the division rule. Exam tip: when dividing powers with the same base, subtract the exponents, not the bases.

Since \(8000=20^3\), the equation \(x^3=20^3\) gives \(x=20\). The cubes of 18, 22, and 24 are not equal to 8000. Exam tip: In such questions, identify the perfect cube and take its cube root directly.

Using the power-of-a-power rule, \\(a^m\\)^n = a^{mn}\\. Therefore, \\( (2^3)^4 = 2^{3×4} = 2^{12} = 4096 \\). Hence, 4096 is correct. Remember that the exponents are multiplied, not added, in this form.

By direct calculation: 31^2 = 961 and 32^2 = 1024, so 961 + 1024 = 1985.
Alternatively use the identity


((n-1)^2 + n^2 = 2n^2 - 2n + 1) with n = 32:
2·32^2 - 2·32 + 1 = 2·1024 - 64 + 1 = 1985.
Common distractor (1975) results from arithmetic slip when squaring or adding. Exam tip: use algebraic identities or split squares (e.g. 32^2 = (30+2)^2) for faster, less error-prone calculation.

First evaluate both powers: 4^5 = 1024 and 2^6 = 64. Therefore, 4^5 - 2^6 = 1024 - 64 = 960, so option A is correct. Exam tip: evaluate the exponents before performing the subtraction.

Since \(37^2=37\times37=1369\), the positive solution of \(y^2=1369\) is \(y=37\). The negative solution \(-37\) is also possible, but the question asks for the positive value. Exam tip: verify a square-root answer by squaring it.

Direct calculation: \(32^2=1024\) and \(33^2=1089\), so the sum is 1024+1089=2113. Quick formula: \(n^2+(n+1)^2=2n^2+2n+1\); with n=32 this gives 2113 as well. Why distractors are wrong: 2103 and 2123 are 10 away and 2119 is 6 away from the correct sum, so they result from simple arithmetic mistakes. Exam tip: use the formula for consecutive integers' squares to compute such sums faster and avoid addition errors.

For division of powers with the same non-zero base, use \(a^m \div a^n=a^{m-n}\). Thus, \(7^5 \div 7^3=7^{5-3}=7^2=49\). Option B incorrectly uses \(7^3\), instead of subtracting the exponents. Exam tip: when dividing powers with the same base, subtract the exponents; do not multiply them.

When powers with the same base are multiplied, their exponents are added: \(2^6 \times 2^5=2^{6+5}=2^{11}\). Since \(2^{11}=2048\), option A is correct. Option B corresponds to \(2^{10}\), while option C corresponds to \(2^{12}\). Exam tip: remember the rule \(a^m \times a^n=a^{m+n}\) for multiplying powers with the same base.

When powers with the same non-zero base are divided, their exponents are subtracted: \(a^m \div a^n=a^{m-n}\). Thus, \(3^8 \div 3^4=3^{8-4}=3^4=81\), so option B is correct. Exam tip: for division with the same base, subtract the exponents rather than multiplying them.

Using the power-of-a-power rule \\(a^m\\)^n = a^{mn}, we get \\(4^3\\)^2 = 4^{3×2} = 4^6 = 4096. Therefore, option C is correct. Exam tip: when a power is raised to another power, multiply the exponents rather than adding them.

When powers with the same base are divided, their exponents are subtracted: \(a^m \div a^n=a^{m-n}\). Therefore, \(5^5 \div 5^3=5^{5-3}=5^2=25\), so option D is correct. Exam tip: For division with the same base, subtract the exponents rather than multiplying them.