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NDA Maths · Binary Numbers

Binary Arithmetic — Addition, Division & Algebraic Identities

Add, subtract and divide in base 2 — or, for almost every NDA question, convert to decimal, do the arithmetic you already know, and convert back. The same subtopic also hides a few algebra-identity questions whose only twist is that the numbers arrive in binary.

All Binary Numbers notesNDA Maths strategy

Why this matters

This is the chapter's largest and hardest pocket: 7 PYQs, 3 of them HARD. Two genres recur. The first is straight arithmetic — sums and divisions, often with unknown bits (p, q, r or x, y) to pin down. The second is an algebra identity (a cube-sum or difference-of-cubes relation) where recognising the structure beats grinding the numbers. Convert-first handles the arithmetic; the identity questions reward spotting x = y + z or a²+b²+c² before you compute anything.

Concept 1 of 3

Binary Addition, Subtraction & Unknown-Digit Puzzles

Intuition

You can add binary numbers bit-by-bit with carries (1 + 1 = 10, carry the 1), exactly like decimal column addition. But for NDA the safe route is almost always: convert each number to decimal, add or subtract there, then convert the result back. Unknown-bit puzzles become a small equation once everything is decimal.

Definition

Binary addition rules (per column, right to left): 0+0=00+0=0, 0+1=10+1=1, 1+1=101+1=10 (write 0, carry 1), 1+1+1=111+1+1=11 (write 1, carry 1). Convert-first method (recommended): turn each binary into decimal, add or subtract in decimal, then convert the answer back to binary. Unknown-digit puzzles: when bits like p,q,rp, q, r or x,yx, y are unknown, write each binary in decimal keeping the unknowns as variables (e.g. (1p101)2=16+8p+5(1p101)_2 = 16 + 8p + 5), form the equation the problem states, and solve — remembering every unknown bit is itself 00 or 11.

Binary addition carry rule

1+1=(10)2,1+1+1=(11)21 + 1 = (10)_2,\qquad 1 + 1 + 1 = (11)_2

Worked example

Find (10110)2+(1101)2(10110)_2 + (1101)_2, giving the answer in binary.
  1. Convert: (10110)2=16+4+2=22(10110)_2 = 16 + 4 + 2 = 22 and (1101)2=8+4+1=13(1101)_2 = 8 + 4 + 1 = 13.
  2. Add in decimal: 22+13=3522 + 13 = 35.
  3. Convert back: 35=32+2+1=(100011)235 = 32 + 2 + 1 = (100011)_2.
Answer:(10110)2+(1101)2=(100011)2(10110)_2 + (1101)_2 = (100011)_2.
Practice this conceptself-check · 4 quick reps

Try it yourself

In (11p1)2+(101)2=(10100)2(11p1)_2 + (101)_2 = (10100)_2, where pp is a binary digit, find pp.

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    Find (101)2+(11)2(101)_2 + (11)_2 in binary.
  2. 2.
    Find (1110)2−(101)2(1110)_2 - (101)_2 in binary.
  3. 3.
    Find (1111)2+(1)2(1111)_2 + (1)_2 in binary.
  4. 4.
    If (1x0)2=6(1x0)_2 = 6, what is the bit xx?

From the bank · past-year question

Example 1Binary NumbersMODERATE
What is the sum of the binary numbers (101101101)2(101101101)_2 and (100011)2(100011)_2?

[Q41 · Sep · 2025]

Every unknown is a BIT — only 0 or 1 is allowed

When you solve for p,q,r,x,yp, q, r, x, y, the value must be 0 or 1. A solution like p=2p = 2 is impossible in base 2 — it means you mis-set the place values. Check each unknown lands in {0,1}\{0, 1\} before choosing an option.

Convert the FINAL answer back to binary

If the question gives the numbers in binary, the options are usually binary too. Doing the addition in decimal is fine — but don't forget the last step of converting your decimal total back to base 2.
Drill 2 more on binary addition, subtraction & unknown-digit puzzles

Concept 2 of 3

Binary Division — Quotient and Remainder

Intuition

Dividing one binary number by another is just ordinary integer division in disguise. Convert both to decimal, divide to get a whole-number quotient and a remainder, then convert each of those back to binary if the options ask for it.

Definition

To compute (A)2÷(B)2(A)_2 \div (B)_2:

  • Convert AA and BB to decimal.
  • Do integer division: find the whole quotient QQ and remainder RR with A=BQ+RA = BQ + R and 0≤R<B0 \le R < B.
  • Convert QQ (and RR, if asked) back to binary.

When the division is exact the remainder is 00; otherwise the remainder is strictly less than the divisor — the same rule as decimal long division.

Division identity

A=B Q+R,0≤R<BA = B\,Q + R,\qquad 0 \le R < B

Worked example

Find the quotient and remainder of (11011)2÷(100)2(11011)_2 \div (100)_2, in binary.
  1. Convert: (11011)2=16+8+2+1=27(11011)_2 = 16 + 8 + 2 + 1 = 27 and (100)2=4(100)_2 = 4.
  2. Integer division: 27=4×6+327 = 4 \times 6 + 3, so quotient =6= 6, remainder =3= 3.
  3. Convert back: 6=(110)26 = (110)_2, 3=(11)23 = (11)_2.
Answer:Quotient (110)2(110)_2, remainder (11)2(11)_2.
Practice this conceptself-check · 4 quick reps

Try it yourself

What is (100100)2÷(110)2(100100)_2 \div (110)_2?

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    Find (1010)2÷(10)2(1010)_2 \div (10)_2.
  2. 2.
    Quotient and remainder of (1011)2÷(11)2(1011)_2 \div (11)_2?
  3. 3.
    Find (110000)2÷(1000)2(110000)_2 \div (1000)_2.
  4. 4.
    Is (1111)2÷(101)2(1111)_2 \div (101)_2 exact?

From the bank · past-year question

Example 2Binary NumbersMODERATE
What is (1110011)2÷(10111)2(1110011)_2\div(10111)_2 equal to?

[Q53 · Apr · 2022]

Quotient and remainder are usually asked in BINARY

After dividing in decimal you have two numbers to convert back — both the quotient and the remainder. Reading the remainder option in decimal (e.g. picking 4 instead of (100)2(100)_2) is the standard slip.
Drill 1 more on binary division — quotient and remainder

Concept 3 of 3

Algebraic Identities with Binary-Given Values

Intuition

Some of the hardest-looking questions in this chapter are really algebra: a cube-sum or difference-of-cubes identity, where the only reason it's filed under Binary is that the given numbers are written in base 2. Convert the binaries to decimal, then let a standard identity do the work instead of cubing huge numbers.

Definition

The recurring identities (after converting every binary to decimal):

  • Sum of cubes: x3+y3=(x+y)(x2−xy+y2)x^3 + y^3 = (x+y)(x^2 - xy + y^2). Given x3+y3x^3+y^3 and x+yx+y, you get x2−xy+y2x^2 - xy + y^2 by dividing — and note (x−y)2+xy=x2−xy+y2(x-y)^2 + xy = x^2 - xy + y^2, so the same quantity answers both.
  • **The a=b+ca = b + c identity:** if a=b+ca = b + c then a3−b3−c3=3bc aa^3 - b^3 - c^3 = 3bc\,a, so a3−b3−c3−3abc=0a^3 - b^3 - c^3 - 3abc = 0. Whenever three given values satisfy one equals the sum of the other two, this expression collapses to zero.

Spot the relationship between the converted values before computing — it usually removes all the heavy arithmetic.

Key cube identities

x3+y3=(x+y)(x2−xy+y2);a=b+c ⇒ a3−b3−c3−3abc=0x^3 + y^3 = (x+y)(x^2 - xy + y^2);\qquad a = b + c \ \Rightarrow\ a^3 - b^3 - c^3 - 3abc = 0

Worked example

Let a=(1010)2, b=(110)2, c=(100)2a = (1010)_2,\ b = (110)_2,\ c = (100)_2. Evaluate a3−b3−c3−3abca^3 - b^3 - c^3 - 3abc.
  1. Convert: a=10, b=6, c=4a = 10,\ b = 6,\ c = 4.
  2. Check the relationship: b+c=6+4=10=ab + c = 6 + 4 = 10 = a, so a=b+ca = b + c.
  3. By the identity a=b+c⇒a3−b3−c3=3abca = b + c \Rightarrow a^3 - b^3 - c^3 = 3abc, so a3−b3−c3−3abc=0a^3 - b^3 - c^3 - 3abc = 0.
Answer:00.
Practice this conceptself-check · 4 quick reps

Try it yourself

If x+y=(1010)2x + y = (1010)_2 and xy=(10101)2xy = (10101)_2, find x2+y2x^2 + y^2 in decimal.

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    If x+y=5x + y = 5 and xy=6xy = 6, find (x−y)2(x - y)^2.
  2. 2.
    If a=b+ca = b + c, what is a3−b3−c3−3abca^3 - b^3 - c^3 - 3abc?
  3. 3.
    Factor x3+y3x^3 + y^3.
  4. 4.
    If x3+y3=35x^3 + y^3 = 35 and x+y=5x + y = 5, find x2−xy+y2x^2 - xy + y^2.

From the bank · past-year question

Example 3Binary NumbersHARD
If x3+y3=(100010111)2x^3+y^3=(100010111)_2 and x+y=(11111)2x+y=(11111)_2, then what is (x−y)2+xy(x-y)^2+xy equal to?

[Q54 · Apr · 2022]

Spot the identity before cubing anything

Cubing two-digit numbers by hand is slow and error-prone. The questions are engineered so that, after converting, either a=b+ca = b + c or a sum/difference-of-cubes factoring applies. Look for that structure first — the brute-force route is the trap.

(x − y)² + xy equals x² − xy + y²

Expanding, (x−y)2+xy=x2−2xy+y2+xy=x2−xy+y2(x-y)^2 + xy = x^2 - 2xy + y^2 + xy = x^2 - xy + y^2. So a question asking for (x−y)2+xy(x-y)^2 + xy is secretly asking for the cube-sum cofactor x2−xy+y2x^2 - xy + y^2 — recognise them as the same target.
Drill 1 more on algebraic identities with binary-given values

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 (3)

Watch out for (5)

Mastery check — 4 interleaved questions

Try each one before clicking. Questions are interleaved across the concepts above, not grouped — interleaving sharpens transfer.

Example 1Binary NumbersHARD
The sum of the binary numbers (11011)2(11011)_2, (10110110)2(10110110)_2 and (10011x0y)2(10011x0y)_2 is the binary number (101101101)2(101101101)_2. What are the values of x and y?

[Q29 · Sep · 2018]

Example 2Binary NumbersEASY
The remainder and the quotient of the binary division (101110)2÷(110)2(101110)_2 \div (110)_2 are respectively

[Q2 · Sep · 2017]

Example 3Binary NumbersMODERATE
If x=(1111)2, y=(1001)2x=(1111)_2,\ y=(1001)_2 and z=(110)2z=(110)_2, then what is x3−y3−z3−3xyzx^3-y^3-z^3-3xyz equal to?

[Q7 · Apr · 2025]

Example 4Binary NumbersHARD
In the binary equation (1p101)2+(10q1)2=(100r00)2(1p101)_2 + (10q1)_2 = (100r00)_2, where pp, qq and rr are binary digits, what are the possible values of pp, qq and rr respectively?

[Q24 · Apr · 2017]

Drill every past-year question on this subtopic

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

Drill the 7 questions