Number Systems, Binary Arithmetic and Logarithms
Two short topics that Paper I names and most other entrance papers do not, which makes them the cheapest marks in the paper: an hour spent here is worth more than an hour spent on calculus, because the questions are mechanical and the competition has usually skipped them. Number systems are pure procedure — divide by two and read the remainders upward to go into binary, multiply by place values to come back out — and binary arithmetic has exactly four addition facts to know. Logarithms are three laws and one change-of-base formula, and the single question type they generate most often is counting the digits of a large power, which needs nothing but log₁₀ 2 = 0.3010 and log₁₀ 3 = 0.4771 held in memory. Both topics reward being able to check your answer: a binary conversion can be converted back in seconds, and a logarithm can be sanity-checked by asking which two integers the answer must lie between.
Logarithms — Three Laws and a Change of Base
log_a x = y means a^y = x. The base must be positive and not 1, and x must be positive — the logarithm of zero or of a negative number is undefined, and a question whose four options include log 0 is usually asking exactly that. Three laws cover the algebra: log(mn) = log m + log n, log(m/n) = log m − log n, and log(mⁿ) = n log m. Two more identities close the topic: log_a 1 = 0 for every valid base, log_a a = 1, and a^(log_a x) = x, because the two operations undo one another.
- Change of base: log_a x = log_b x / log_b a. The immediate consequence, asked directly, is that log_a b × log_b a = 1.
- Exponents in the base move out as a ratio: log_(aⁿ)(x^m) = (m/n)·log_a x. So log₄ x is half of log₂ x, which is the step that turns a two-base equation into a one-base equation.
- A negative logarithm is a fraction inside. log₃(1/81) = log₃(3⁻⁴) = −4. Nothing has gone wrong: a logarithm is negative exactly when its argument lies between 0 and 1.
🎯 Counting the digits of a large power
The number of digits in N is ⌊log₁₀ N⌋ + 1. So 2⁵⁰ has ⌊50 × 0.3010⌋ + 1 = ⌊15.05⌋ + 1 = 16 digits, and 3²⁰ has ⌊20 × 0.4771⌋ + 1 = ⌊9.542⌋ + 1 = 10 digits. This is the standard question type of the topic, and it needs only log₁₀ 2 = 0.3010 and log₁₀ 3 = 0.4771 memorised. Note the floor: forgetting it, or adding 1 before flooring, is the one slip that produces a confident off-by-one answer.Key takeaways
- Both topics are in the NDA syllabus and in almost no competing paper, so they are the cheapest marks available — short, mechanical, and usually unprepared by the competition.
- Decimal to binary: divide by 2 and read the remainders upward. Binary to decimal: multiply by place values and add. Always convert back to check.
- 2ⁿ − 1 is n ones in binary: 7 = 111, 15 = 1111, 31 = 11111, 255 = 11111111. Recognising the pattern replaces the whole conversion.
- Binary addition is four facts: 0+0 = 0, 0+1 = 1, 1+1 = 10, 1+1+1 = 11. Check the sum in decimal afterwards.
- The logarithm of zero or of a negative number is undefined. An option offering log 0 is there to be rejected, not computed.
- log_a b × log_b a = 1, and an exponent in the base comes out as a ratio: log₄ x is half of log₂ x. That step collapses a two-base equation to one base.
- Digits in N = ⌊log₁₀ N⌋ + 1. With log 2 = 0.3010 and log 3 = 0.4771 memorised, every version of this question is one multiplication and a floor.
Practice questions (8)
Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.
The binary equivalent of the decimal number 45 is:
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Answer: A — 101101
Divide by 2 and read the remainders upward: 45→22 r1, 22→11 r0, 11→5 r1, 5→2 r1, 2→1 r0, 1→0 r1, giving 101101. Check by converting back: 32 + 8 + 4 + 1 = 45. Option B is 110101 = 53 — that is the same six digits read in the wrong direction, which is the error this question is built to catch. Option C is 43 and option D is 57.The decimal value of the binary number 110101 is:
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Answer: C — 53
Multiply each digit by its place value, right to left: 1 + 0 + 4 + 0 + 16 + 32 = 53. Option A, 45, is 101101 — this question and the previous one are deliberately the same digits in opposite orders, because a candidate who reads remainders the wrong way round will produce exactly this pair of answers and have no way of noticing.The sum of the binary numbers 1011 and 110 is:
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Answer: B — 10001
Add column by column with 1+1 = 10 (write 0, carry 1): 1011 + 0110 gives 1+0 = 1; 1+1 = 10 so write 0 carry 1; 0+1+1 = 10 so write 0 carry 1; 1+0+1 = 10 so write 0 carry 1; and the final carry gives the leading 1 — 10001. Verify in decimal: 11 + 6 = 17, and 17 = 16 + 1 = 10001. Option A, 1111, is 15 and comes from adding the columns without carrying at all.Which of the following decimal numbers is written as all ones in binary?
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Answer: B — 31
A number written as n ones in binary is 2ⁿ − 1, because 1 + 2 + 4 + … + 2ⁿ⁻¹ = 2ⁿ − 1. So 31 = 2⁵ − 1 = 11111. The neighbours are instructive: 32 = 2⁵ is 100000, a single one followed by five zeros; 30 = 11110 and 33 = 100001. Learning the pattern in both directions — one less than a power of two is all ones, a power of two itself is one followed by zeros — answers a whole family of questions without any division.The value of log₂ 8 + log₃ 9 is:
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Answer: B — 5
Read each logarithm as a question about an exponent. log₂ 8 asks "2 to what power is 8?" — the answer is 3, since 8 = 2³. log₃ 9 asks "3 to what power is 9?" — the answer is 2, since 9 = 3². So the sum is 5. Option D, 17, comes from adding the arguments (8 + 9) instead of the logarithms, which is the commonest error in the topic: log m + log n equals log(mn) only when the bases are the same, and here they are not.The value of log₃ (1/81) is:
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Answer: B — −4
1/81 = 3⁻⁴, so log₃(3⁻⁴) = −4. A logarithm is negative exactly when its argument lies strictly between 0 and 1, and 1/81 does — so the minus sign is the expected result, not an error to be corrected. Option D is wrong for an important reason: 1/81 is positive, and only a zero or negative argument makes a logarithm undefined. Confusing "a fraction" with "not allowed" is the trap this question sets.Given log₁₀ 2 = 0.3010, the number of digits in 2⁵⁰ is:
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Answer: B — 16
The number of digits in N is ⌊log₁₀ N⌋ + 1. Here log₁₀(2⁵⁰) = 50 × 0.3010 = 15.05, so the count is ⌊15.05⌋ + 1 = 16. Option A drops the +1 and reports the characteristic instead of the digit count; option C rounds 15.05 up to 16 before adding 1, which double-counts. The formula is worth stating carefully because both slips give an answer that sits right next to the correct one, and nothing in the question flags either.Which of the following is NOT defined?
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Answer: B — log₁₀ 0
A logarithm needs a positive argument, so log₁₀ 0 is undefined — no power of 10 gives 0. The other three all exist: log₁₀ 1 = 0 (any base to the power 0 is 1); log₁₀ 0.1 = −1, negative but perfectly defined; and log₀.₁ 10 = −1 as well, because 0.1 is a legal base — a base must be positive and not equal to 1, and 0.1 satisfies both. Options C and D are offered because candidates learn "positive base, positive argument" and then reject anything that looks small or fractional.