NDA Maths · Binary Numbers

Binary Representation & Number Theory

A small grab-bag the NDA files under Binary Numbers: representing a decimal number in binary (and counting its bits), plus a couple of pure number-theory recall items — modular remainders by cycling, and the sum-of-odd-numbers perfect-square fact.

All Binary Numbers notes NDA Maths strategy

Why this matters

Three PYQs sit here. One is a direct decimal-to-binary representation; the other two are number-theory one-liners (a remainder-by-cycling and a perfect-square recognition) that happen to be grouped with binary in the syllabus. They reward two reflexes you can memorise: powers repeat in a short cycle modulo a number, and 1 + 3 + 5 + ... + (2n − 1) = n².

Concept 1 of 2

Representing a Number in Binary & Counting Its Bits

Intuition

Representing a decimal number in binary is the same repeated-division skill from Subtopic 1, but here the question often cares about HOW MANY bits the answer needs — which is governed by where the number sits between consecutive powers of 2.

Definition

A decimal number $N \ge 1$ needs exactly $k$ bits in binary when it lies in the range

2^{k-1} \le N \le 2^k - 1.

Equivalently, the number of bits is one more than the highest power of 2 not exceeding

N

. For example

1011

lies between

2^9 = 512

and

2^{10} = 1024

, so its binary form uses 10 bits; converting by repeated division gives

1011 = (1111110011)_2

A number that is exactly $2^k$ is the smallest $(k+1)$ -bit number: a 1 followed by $k$ zeros.
A number that is exactly $2^k - 1$ is the largest $k$ -bit number: $k$ ones.

Bit count of N

2^{k-1} \le N \le 2^k - 1 \ \Longrightarrow\ N \text{ uses } k \text{ bits}

Worked example

How many bits are needed to write

100_{10}

in binary, and what is that binary form?

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 1Binary NumbersEASY

What is the binary number equivalent to decimal number 1011?

[Q3 · Apr · 2023]

Watch the digit COUNT in the options

Representation questions often offer distractors with the wrong number of bits (one too few or too many). Bracket

N

between consecutive powers of 2 first to know how many bits the correct answer must have, then convert.

Concept 2 of 2

Number-Theory One-Liners — Remainder Cycles & Sum of Odd Numbers

Intuition

These two questions aren't about binary at all — they're number-theory facts the syllabus groups here. The first uses the fact that powers repeat in a short cycle when you take remainders. The second uses the clean identity that adding the first n odd numbers always gives a perfect square, n².

Definition

Remainder by cycling: the remainders of $a^1, a^2, a^3, \ldots$ modulo a fixed number repeat with some period $L$ . Find the cycle by computing remainders until they repeat, then reduce the exponent modulo $L$ : if the exponent leaves remainder $t$ on division by $L$ , then $a^{\text{exp}}$ has the same remainder as $a^t$ . Sum of the first n odd numbers:

1 + 3 + 5 + \cdots + (2n - 1) = n^2.

So if a sum of consecutive odd numbers from 1 equals some value

S

, the number of terms is

n = \sqrt{S}

— and

S

must be a perfect square. (A neat NDA case:

\sqrt{12345678987654321} = 111111111

Sum of first n odd numbers

1 + 3 + 5 + \cdots + (2n - 1) = n^2

Worked example

What is the remainder when

3^{100}

is divided by 7?

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 2Binary NumbersMODERATE

What is the remainder when

5^{99}

is divided by 13?

[Q9 · Sep · 2025]

Reduce the exponent by the CYCLE length, not the modulus

For

5^{99} \bmod 13

, the powers cycle with period 4 (not 13). Reduce

99

modulo 4, not modulo 13. Find the actual cycle length first by listing remainders until one repeats.

Sum of odd numbers is a PERFECT SQUARE

If a sum of

1 + 3 + 5 + \cdots

is given, the term count is

\sqrt{\text{sum}}

. Recognising a value like

12345678987654321

111111111^2

is the intended shortcut — don't try to add the series term by term.

Drill 1 more on number-theory one-liners — remainder cycles & sum of odd numbers

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Representing a Number in Binary & Counting Its Bits
Bit count of N
$2^{k-1} \le N \le 2^k - 1 \ \Longrightarrow\ N \text{ uses } k \text{ bits}$
Number-Theory One-Liners — Remainder Cycles & Sum of Odd Numbers
Sum of first n odd numbers
$1 + 3 + 5 + \cdots + (2n - 1) = n^2$

Watch out for (3)

Watch the digit COUNT in the options
→ Representing a Number in Binary & Counting Its Bits
Reduce the exponent by the CYCLE length, not the modulus
→ Number-Theory One-Liners — Remainder Cycles & Sum of Odd Numbers
Sum of odd numbers is a PERFECT SQUARE
→ Number-Theory One-Liners — Remainder Cycles & Sum of Odd Numbers

Drill every past-year question on this subtopic

3 questions from the bank — paginated, with cart and Word-export support.

Drill the 3 questions