Quick Summary: Powers, exponents, surds, logarithms and the number system contribute 1 to 3 marks in GATE General Aptitude 2027 across every paper (DA, RA, CS, ME, EE, CE, EC). The questions test a handful of laws: exponent rules, log identities, unit-digit cycles, divisibility, remainders and counting factors. This guide covers every rule GATE uses, 8 solved previous-year and GATE-style questions, a 5-question practice set, a one-page formula sheet and the traps that cost a mark you had already earned.
📋 Table of Contents
Why this topic matters in GATE GA
This is the chapter where GATE rewards people who know rules and punishes people who compute. Nobody can evaluate 21717 in the exam hall, but anybody who knows that unit digits of powers repeat in cycles of four can find its last digit in ten seconds. The same applies to logarithms: a question that looks like it needs a calculator usually collapses to one identity.
Across DA, RA, CS, ME, EE, CE and EC the GA section is common, so this chapter is worth the same to everybody. The questions come in two flavours: a 1-mark MCQ on an exponent or log identity, and a 2-mark NAT on unit digits, remainders or the number of factors. Both are quick if the rule is in your head and impossible if it is not.
Here is roughly how much this cluster has contributed. Aspirants who want 15/15 in GA should be able to finish every question in this chapter under 60 seconds.
| Year | Marks from this topic (approximate, across papers) | Typical question type |
|---|---|---|
| 2019 | 1 to 2 | Exponent equation, number of integers in a range |
| 2020 | 1 to 2 | Logarithm identity, surd simplification |
| 2021 | 1 to 3 | Remainder, divisibility |
| 2022 | 1 to 2 | Exponent comparison, unit digit |
| 2023 | 1 to 2 | Powers of 2 in a sequence, factor counting |
| 2024 | 1 to 3 | Logarithm equation, HCF and LCM |
| 2025 | 1 to 2 | Exponential growth, last digit |
| 2026 | 1 to 2 | Surds, perfect squares in a range |
What the official GA syllabus says
“Powers, exponents and logarithms” is the exact phrase. The number-system questions (divisibility, remainders, factors) come under “numerical computation and estimation”. The complete syllabus breakdown and weightage is in the GATE General Aptitude 2027 hub. The previous post in this series covered Ratio, Proportion, Percentage and Averages.
Laws of exponents
For a positive base a and any real m, n:
- am × an = am+n; am ÷ an = am−n
- (am)n = amn; (ab)n = anbn; (a/b)n = an/bn
- a0 = 1 (a ≠ 0); a−n = 1/an; a1/n = n√a; am/n = (n√a)m
- If am = an with a ≠ 0, 1, −1, then m = n. If an = bn with n odd, then a = b.
The most common GATE trick is to write both sides with the same base. 4x = 8x−1 becomes 22x = 23x−3, so 2x = 3x − 3 and x = 3.
Comparing powers: to compare 2300, 3200 and 5100, take the 100th root: 23 = 8, 32 = 9, 51 = 5. So 3200 is largest, 5100 is smallest.
Solution: 3x(32 − 1) = 72, so 3x × 8 = 72 and 3x = 9. Hence x = 2. (Check: 34 − 32 = 81 − 9 = 72.)
Solution: Number of doublings = 24/3 = 8. Cells = 500 × 28 = 500 × 256 = 1,28,000.
Surds and rationalisation
A surd is an irrational root such as √2 or 3√5. Rules:
- √a × √b = √(ab); √a / √b = √(a/b); but √a + √b ≠ √(a + b).
- Rationalise 1/(√a + √b) by multiplying numerator and denominator by (√a − √b), giving (√a − √b)/(a − b).
- Nested surds: √(a + 2√b) = √x + √y where x + y = a and xy = b. For √(7 + 2√12): x + y = 7, xy = 12, so x = 4, y = 3, and the answer is 2 + √3.
- Useful values: √2 ≈ 1.414, √3 ≈ 1.732, √5 ≈ 2.236, √10 ≈ 3.162.
Solution: Rationalise: x = (√5 + 2)/(5 − 4) = √5 + 2. Then 1/x = √5 − 2. Sum = 2√5, approximately 4.472.
Logarithms
logax = y means ay = x, with a > 0, a ≠ 1, x > 0. Everything else follows from the exponent laws.
- log(mn) = log m + log n; log(m/n) = log m − log n; log mk = k log m
- logaa = 1; loga1 = 0; alogax = x
- Change of base: logax = logbx / logba; in particular logab × logba = 1
- logakx = (1/k) logax
- Values to remember: log102 ≈ 0.3010, log103 ≈ 0.4771, log105 = 1 − log102 ≈ 0.6990, log107 ≈ 0.8451
Number of digits: an integer N has floor(log10N) + 1 digits. So 2100 has floor(100 × 0.3010) + 1 = 30 + 1 = 31 digits.
Solution: log4x = (1/2) log2x, so (3/2) log2x = 6 and log2x = 4. Hence x = 16. (Check: log216 = 4, log416 = 2, sum 6.)
Number system: divisibility, unit digits, remainders, factors
Divisibility rules
- By 2, 5, 10: look at the last digit. By 4: last two digits divisible by 4. By 8: last three digits divisible by 8.
- By 3 and 9: digit sum divisible by 3 or 9.
- By 11: (sum of digits in odd places) − (sum of digits in even places) is 0 or a multiple of 11.
- By 6: divisible by both 2 and 3. By 12: by both 3 and 4. By 7: double the last digit and subtract from the rest; repeat.
Unit digit of a power
Only the unit digit of the base matters. Digits 0, 1, 5, 6 never change. Digit 4 alternates 4, 6 (odd power gives 4, even gives 6); digit 9 alternates 9, 1. Digits 2, 3, 7, 8 repeat in cycles of four:
| Unit digit of base | Power 1 | Power 2 | Power 3 | Power 4 |
|---|---|---|---|---|
| 2 | 2 | 4 | 8 | 6 |
| 3 | 3 | 9 | 7 | 1 |
| 7 | 7 | 9 | 3 | 1 |
| 8 | 8 | 4 | 2 | 6 |
Divide the exponent by 4. Remainder 1, 2, 3 picks column 1, 2, 3; remainder 0 picks column 4. Unit digit of 72027: 2027 = 4 × 506 + 3, so column 3, answer 3.
Remainders
Remainders respect addition and multiplication. To find 2100 mod 7: 23 = 8 ≡ 1 (mod 7), and 100 = 3 × 33 + 1, so 2100 ≡ (23)33 × 2 ≡ 1 × 2 = 2. Find a power of the base that leaves remainder 1 (or −1) and the rest is arithmetic.
Factors, HCF and LCM
If N = pa qb rc (prime factorisation), then the number of factors is (a + 1)(b + 1)(c + 1) and the sum of factors is (1 + p + … + pa)(1 + q + … + qb)(1 + r + … + rc). For two numbers, HCF × LCM = product of the numbers.
Solution: 360 = 23 × 32 × 5. Factors = (3 + 1)(2 + 1)(1 + 1) = 4 × 3 × 2 = 24. Odd factors use no 2: (2 + 1)(1 + 1) = 6 (they are 1, 3, 5, 9, 15, 45).
The GATE Aptitude Test Series has a dedicated topic test on exponents, logs and number system with 40 timed questions of exactly this type.
Shortcuts and traps
❌ Writing √(a + b) = √a + √b or log(a + b) = log a + log b. Neither is true.
❌ Treating am × bm as (ab)2m. It is (ab)m.
❌ Using exponent remainder 0 as “column 0” in the unit-digit table. Remainder 0 means the fourth column.
❌ Counting 1 and N themselves out of the factor count. The formula (a + 1)(b + 1) includes both.
❌ Forgetting that log is undefined for zero and negative arguments; a “solution” that makes x ≤ 0 must be rejected.
❌ Confusing loga(x2) = 2 logax with (logax)2.
Solved previous-year and GATE-style questions
Solution: Only unit digits matter: 17 ends in 1. 29: 9 = 4 × 2 + 1, column 1, ends in 2. 311: 11 = 4 × 2 + 3, column 3, ends in 7. 413: odd power, ends in 4. Sum of unit digits = 1 + 2 + 7 + 4 = 14. Answer: (B) 4.
Solution: Set each equal to k: log P = k, log Q = 2k, log R = 3k. Then log(Q2) = 4k and log(PR) = k + 3k = 4k. Answer: (A) Q2 = PR.
Solution: 1259 + 2062 = 3321, so (1.001)3321 = 3.52 × 7.85 = 27.632. Answer: (D) 27.64.
Solution: The three relations say qa = r, rb = s, sc = q. Substitute: s = rb = qab, and q = sc = qabc. Hence abc = 1. Answer: (C) 1.
Solution: 2100 = 22 × 3 × 52 × 7. Number of divisors = (2 + 1)(1 + 1)(2 + 1)(1 + 1) = 3 × 2 × 3 × 2 = 36.
Solution: x−1/3 = 5/7, so x1/3 = 7/5 and x = (7/5)3 = 343/125.
Solution: 36 = 729 = 7 × 104 + 1, so 36 ≡ 1 (mod 7). 100 = 6 × 16 + 4, so 3100 ≡ 34 = 81 = 7 × 11 + 4. Remainder: 4.
Solution: Digit sum = 7 + x + 5 + 4 + 2 + 3 = 21 + x. For divisibility by 9 the sum must be 27 (the only multiple of 9 reachable with a digit 0 to 9), so x = 6.
★ PiyushAI Test Series
Best GATE Aptitude Test Series 2027
General Aptitude is 15 marks in every GATE paper and the easiest 15 marks to lose to silly mistakes. The PiyushAI Aptitude Test Series by Piyush Wairale (IIT Madras) gives you topic-wise tests on exactly this chapter, full-length GA mocks in the real GATE interface, detailed video and text solutions, and All-India rank analysis.
Practice set
Attempt these in 6 minutes before opening the key.
- If 2x = 4y = 8z and xyz = 288, then x + y + z is (A) 22 (B) 24 (C) 11 (D) 36
- The unit digit of 795 − 358 is (A) 0 (B) 4 (C) 6 (D) 2
- If log102 = 0.3010, the number of digits in 520 is (A) 13 (B) 14 (C) 15 (D) 20
- The HCF of two numbers is 12 and their LCM is 180. If one number is 36, the other is (A) 48 (B) 60 (C) 72 (D) 90
- The value of √(11 + 2√30) is (A) √5 + √6 (B) √10 + 1 (C) √3 + 2√2 (D) 2 + √7
Show answer key
- (A) x = 2y = 3z = 6t gives x = 6t, y = 3t, z = 2t; xyz = 36t3 = 288, t3 = 8, t = 2; sum = 11t = 22.
- (B) 795: 95 = 4 × 23 + 3, unit digit 3. 358: 58 = 4 × 14 + 2, unit digit 9. 3 − 9 means borrow: 13 − 9 = 4.
- (B) log 520 = 20(1 − 0.3010) = 13.98; digits = 13 + 1 = 14.
- (B) Other number = HCF × LCM / 36 = 12 × 180/36 = 60.
- (A) x + y = 11, xy = 30, so x = 6, y = 5: √6 + √5. Check: 6 + 5 + 2√30 = 11 + 2√30.
One-page formula sheet
2. am/n = n√(am). Compare powers by taking a common root.
3. √a√b = √(ab). 1/(√a + √b) = (√a − √b)/(a − b).
4. √(a + 2√b) = √x + √y with x + y = a, xy = b.
5. log(mn) = log m + log n; log(m/n) = log m − log n; log mk = k log m.
6. logax = log x / log a; logab × logba = 1; alogax = x.
7. log102 = 0.3010, log103 = 0.4771, log105 = 0.6990, log107 = 0.8451. Digits of N = floor(log10N) + 1.
8. Unit-digit cycles: 2, 3, 7, 8 repeat every 4; 4 and 9 repeat every 2; 0, 1, 5, 6 are constant.
9. Divisibility: 3 and 9 by digit sum; 4 by last two digits; 8 by last three; 11 by alternating sum.
10. Factors of paqbrc = (a + 1)(b + 1)(c + 1). HCF × LCM = product of two numbers.
11. Remainders: find a power of the base that is ≡ 1 or −1, then reduce the exponent.
12. Perfect squares up to N: floor(√N). Perfect cubes up to N: floor(3√N).
Common mistakes
❌ Solving ax = bx as a = b without checking x = 0.
❌ Taking log of both sides of an equation that has a sum inside the log.
❌ Reading “number of digits” as log N rather than floor(log N) + 1.
❌ Forgetting to borrow when subtracting unit digits (3 − 9 gives 4, not −6).
❌ Using the divisibility rule for 11 with the wrong sign order and then not taking the absolute value.
FAQs
How many questions from powers, logs and number system come in GATE GA?
Usually 1 to 2 questions worth 1 to 3 marks per paper. They are quick, rule-based questions; the mark is lost only by not knowing the rule or by careless arithmetic.
Do I need to memorise log values?
Only log102 = 0.3010 and log103 = 0.4771. Every other common log value (5, 6, 8, 9, 12) follows from these two. GATE sometimes gives the values in the question anyway.
Is a scientific calculator available for these questions?
Yes, the GATE interface has a virtual scientific calculator. But using it for unit-digit or remainder questions is slower than the rule, and for a number like 21717 it will overflow or round. Learn the rules; use the calculator only for the final multiplication.
Is the PiyushAI Aptitude Test Series enough for GA?
Yes. The Best GATE Aptitude Test Series 2027 has a topic test on exponents, logarithms and number system, tests on every other GA chapter, full-length GA mocks in the GATE interface, solved PYQs from 2010 to 2026 and All-India rank analysis. With the free guides on this site, it covers everything the GA section asks.
Are surds really asked in GATE?
Rarely as a standalone question, but surds appear inside geometry answers (√2, √3) and inside exponent questions written with fractional powers. Knowing how to rationalise and simplify nested surds is a five-minute investment that keeps paying.
Stop losing easy marks. Practise this topic under exam pressure.
Previous in series: Ratio, Proportion, Percentage & Averages for GATE Aptitude 2027 | Next in series: Permutation, Combination & Probability for GATE Aptitude 2027
Related guides: GATE General Aptitude 2027 hub | All GATE General Aptitude articles | GATE DA Syllabus 2027 | GATE RA Syllabus 2027
Recent Post
Solve Puzzles, Seating Arrangement, Blood Relations and Direction Sense for GATE General Aptitude 2027: circular and linear arrangement method, family-tree decoding, coordinate method for directions, diagrams, 8 solved PYQs, practice set and rule sheet by Piyush Wairale (IIT Madras).
Crack Analogy, Series and Numerical Relations for GATE General Aptitude 2027: number and letter series methods, word and number analogies, coding-decoding, grid and operator puzzles, 8 solved PYQs, practice set and rule sheet by Piyush Wairale (IIT Madras).
Master Logical Deduction and Induction for GATE General Aptitude 2027: syllogisms with Venn diagrams, if-then statements and the contrapositive, truth-teller puzzles, assumption and weaken questions, 8 solved PYQs, practice set and rule sheet by Piyush Wairale (IIT Madras).
Speed-time-distance, work-time, pipes, clocks and calendars for GATE General Aptitude 2027: average and relative speed, trains and boats, the LCM work method, the clock-angle formula, odd days, 8 solved PYQs and GATE-style questions, a practice set and a formula sheet.
Data Interpretation for GATE General Aptitude 2027: how to read tables, bar graphs, line graphs, pie charts and Venn diagrams fast, with worked diagrams, 8 solved PYQs and GATE-style questions, a practice set and a one-page rule sheet.
Mensuration and geometry for GATE General Aptitude 2027: area, perimeter, volume and surface-area formulas for triangles, circles, sectors, cylinders, cones and spheres, with labelled diagrams, 8 solved PYQs and GATE-style questions, a practice set and a formula sheet.
Learn Daily, Wherever You Are
Free lectures, exam updates, PYQ discussions, and job alerts — delivered through our YouTube channel and Telegram communities.

