Exponents

For the same base, exponents are subtracted when dividing: \(a^m \div a^n = a^{m-n}\). Thus \(5^5 \div 5^2 = 5^{5-2} = 5^3 = 125\). Closest distractors: 25 is \(5^2\) (too small), 625 is \(5^4\) (result of a wrong operation like adding exponents), and 5 is \(5^1\) (over-subtracted). Exam tip: when dividing like bases, subtract the exponents — don’t divide the bases themselves.

Compute directly: \(20^2 = 400\). Use the identity for the nearby square: \(21^2 = (20+1)^2 = 20^2 + 2\times20\times1 + 1^2 = 400 + 40 + 1 = 441\). Adding gives \(400 + 441 = 841\), so the correct answer is 841.
The closest wrong choice (831) typically arises from a slip in calculating \(21^2\) or an arithmetic error when adding. Exam tip: use \((a+b)^2\) or the nearest perfect square method to avoid careless mistakes.

Using the power-of-a-power rule, \((a^m)^n=a^{mn}\). Therefore, \((4^2)^3=4^{2\times3}=4^6=4096\), so option A is correct. In the exam, remember to multiply the exponents; option B can result from incorrectly taking the exponent as 5 instead of 6.

First calculate the squares: 21^2 = 441 and 22^2 = 484. Therefore, 21^2 + 22^2 = 441 + 484 = 925, so option B is correct. In an exam, calculating each power separately before adding helps avoid arithmetic errors.

Compute each power first: \(6^4=6\times6\times6\times6=1296\) and \(3^5=3\times3\times3\times3\times3=243\). Adding gives \(1296+243=1539\). Option B (1536) is a close arithmetic mistake (treating 243 as 240). Option A (1296) is just \(6^4\) alone. Exam tip: evaluate powers separately and then add; checking the units digit often catches small addition errors.

Direct calculation: \(22^2=484\) and \(23^2=529\). Adding gives \(484+529=1013\).
Alternate quick method (identity): \(a^2+b^2=(a+b)^2-2ab\). So \((22+23)^2-2\times22\times23=45^2-1012=2025-1012=1013\.
Exam tip: Use this identity for fast mental calculation of sums of squares. Option B (1003) is a common slip (off by 10) and therefore incorrect.

When powers with the same base are divided, their exponents are subtracted: \(a^m \div a^n = a^{m-n}\). Therefore, \(7^4 \div 7^1 = 7^{4-1} = 7^3 = 343\). Option B is the value of only \(7^2\), so it is incorrect. Exam tip: In division of powers with the same base, keep the base unchanged and subtract the exponents.

Since (34)^2 = 34 × 34 = 1156, e = ±34. However, the question asks for the positive value, so e = 34. Remember that the positive square root is selected when a positive value is requested; therefore, −34 is not the answer.

Direct method: \(23^2=529\) and \(24^2=576\). Adding gives \(529+576=1105\).
Alternate quick method: use \(a^2+b^2=(a+b)^2-2ab\). Here \((23+24)^2=47^2=2209\) and \(2\times23\times24=1104\), so \(2209-1104=1105\).
Why other options are wrong: 1100 and 1095 result from small arithmetic undercounts; 1115 is an overestimate.
Exam tip: Use the identity \((a+b)^2-2ab\) to avoid computing two large squares separately — it's faster and reduces errors.

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.