Complete Number System notes for SSC CGL, CHSL, CPO and other competitive exams — concepts, formulas, shortcuts, solved examples and the common traps. Every concept used across all 40 chapter tests is explained here from zero.
Number System is one of the highest-scoring topics in SSC Quantitative Aptitude. Questions from this chapter appear in SSC CGL, SSC CHSL, SSC CPO, SSC MTS and most other competitive exams, and they can usually be solved in under a minute once the underlying rule is clear.
These notes explain every concept from the beginning — no prior preparation is assumed. Each rule comes with a worked example, the exam shortcut, and the mistake that most often costs a mark.
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Every number in the exam belongs to one of a few families. Knowing the family tells you what is allowed and what is not.
The family tree, smallest circle first
Each family sits inside the next: N ⊂ W ⊂ Z ⊂ Q ⊂ R.
Which of these is irrational: √16, √20, 0.75, 2/7 ?
√16 = 4, a whole number — rational. 0.75 ends — rational. 2/7 = 0.285714285714… repeats — rational. √20 is not a perfect square, so its root never ends or repeats — irrational.
Rule to carry: the square root of a whole number is rational only when that number is a perfect square.
A fraction in lowest terms p/q gives a terminating decimal only when q is built from 2s and 5s alone.
Both kinds are rational. Terminating and recurring are both allowed; only "never ends AND never repeats" is irrational.
Turning a recurring decimal into a fraction, without algebra
Zero is a whole number and an integer, but not a natural number, and it is even. Also, 0 is rational (0 = 0/1). Questions asking "the smallest natural number" want 1; "the smallest whole number" wants 0.
Face value = the digit itself, wherever it stands.
Place value = the digit × the value of its position — what it is really worth inside the number.
In 617853, going left from the units place: 3 (units, 1), 5 (tens, 10), 8 (hundreds, 100), 7 (thousands, 1000), 1 (ten-thousands, 10000), 6 (lakhs, 100000).
So the face value of 1 is 1, but its place value is 10000.
In 617853, what is the sum of the place values of 1 and 8?
1 sits in the ten-thousands place → 1 × 10000 = 10000.
8 sits in the hundreds place → 8 × 100 = 800.
Sum = 10800. (If you answered 9, you added the face values.)
Moving right from the point the places are tenths (0.1), hundredths (0.01), thousandths (0.001) — each one a tenth of the last.
In 12.345: 3 is tenths (0.3), 4 is hundredths (0.04), 5 is thousandths (0.005).
Parity arithmetic — decides many answers in one second
Use it as a filter: if the answer must be odd, cross out every even option before calculating.
To test N, divide by primes only, and stop once the prime crosses √N.
Why stop there? If N = a × b with both a and b bigger than √N, then a×b would be bigger than N. So one factor must be at or below √N.
Practical bands: below 121 test 2, 3, 5, 7 | below 169 add 11 | below 289 add 13.
Is 197 prime? √197 ≈ 14, so test up to 13.
Odd (not 2) · digit sum 17, not a multiple of 3 · does not end in 0 or 5 · 197 ÷ 7 ≈ 28.1 · 197 ÷ 11 ≈ 17.9 · 197 ÷ 13 ≈ 15.2.
Nothing divides it → 197 is prime.
These four look prime and are not. Learn them by sight:
91 = 7×13 · 119 = 7×17 · 121 = 11² · 133 = 7×19 · 143 = 11×13 · 169 = 13² · 187 = 11×17 · 209 = 11×19 · 221 = 13×17 · 247 = 13×19
Notice the pattern: they are all products of two two-digit-ish primes, so no small factor exposes them. The 11-test and the 13-test are what catch them.
| Range | Primes | Count |
|---|---|---|
| 1–10 | 2, 3, 5, 7 | 4 |
| 11–20 | 11, 13, 17, 19 | 4 |
| 21–40 | 23, 29, 31, 37 | 4 |
| 41–60 | 41, 43, 47, 53, 59 | 5 |
| 61–80 | 61, 67, 71, 73, 79 | 5 |
| 81–100 | 83, 89, 97 | 3 |
Useful counts to remember: 25 primes below 100, 21 more between 100 and 200 (so 46 below 200).
↑ back to topThese save the most time in the whole paper. Learn them until they are automatic.
| Divisor | Test | Example |
|---|---|---|
| 2 | Last digit is 0, 2, 4, 6 or 8 | 4738 ✔ |
| 3 | Digit sum divisible by 3 | 4152 → 4+1+5+2 = 12 ✔ |
| 4 | Last two digits form a multiple of 4 | 7316 → 16 ✔ |
| 5 | Ends in 0 or 5 | 2350 ✔ |
| 6 | Passes the 2-test and the 3-test | 516: even, digit sum 12 ✔ |
| 7 | Double the last digit, subtract from the rest; repeat | 343 → 34 − 6 = 28 ✔ |
| 8 | Last three digits form a multiple of 8 | 17512 → 512 ✔ |
| 9 | Digit sum divisible by 9 | 5346 → 18 ✔ |
| 10 | Ends in 0 | 8970 ✔ |
| 11 | Alternating digit sum (+ − + − from the right) is 0 or a multiple of 11 | 209 → 9 − 0 + 2 = 11 ✔ |
| 12 | Passes 3-test and 4-test | 936 ✔ |
| 25 | Last two digits are 00, 25, 50 or 75 | 3675 ✔ |
| 125 | Last three digits form a multiple of 125 | 4250 ✔ |
Building a rule for any composite divisor
Split the divisor into co-prime parts and test each. 72 = 8 × 9, so test 8 and 9. 36 = 4 × 9. 45 = 5 × 9.
Do not split into parts that share a factor: 24 is not "4 and 6", because 4 and 6 both carry a 2 and the test would pass numbers that fail 24. Use 8 and 3.
Is 936 divisible by 8, 12 and 18?
8: last three digits 936, and 936 ÷ 8 = 117 ✔
12 = 3 × 4: digit sum 18 ✔, last two digits 36 ✔
18 = 2 × 9: even ✔, digit sum 18 ✔
So 936 clears all three. (This is exactly how "greatest three-digit number divisible by 8, 12 and 18" is checked.)
Some expressions are divisible by a fixed number for every value of n. Two facts do most of the work:
So spot the consecutive run hiding inside the expression:
Why is n⁴ − n² always divisible by 12?
Factor: n²(n−1)(n+1).
The 3: among n−1, n, n+1 one must be a multiple of 3.
The 4: if n is even, n² already carries 4. If n is odd, then n−1 and n+1 are two consecutive even numbers, and one of any two consecutive evens is a multiple of 4.
3 and 4 are co-prime, so 3 × 4 = 12 always divides it.
In "always divisible by" questions, test the smallest case first. For n⁴ − n² at n = 2 the value is 12 exactly — which instantly kills 18 and 24. If you had tested n = 3 (72) or n = 4 (240) you would have been tempted by 24.
And if two options are both always true (say 6 and 12), the question must be asking for the largest such number. Read the wording.
This is the single most useful section in the chapter. Roughly one question in five starts here.
Step one, always: prime factorisation. Break the number into primes with powers.
Divide by 2 as long as you can, then 3, then 5, then 7, and so on.
360 → 180 → 90 → 45 (three 2s), 45 → 15 → 5 (two 3s), then 5.
So 360 = 2³ × 3² × 5.
The four formulas. If N = pa × qb × rc …
Why the +1? Because the power of p in a factor may be 0, 1, 2, … up to a — that is a+1 choices, and the 0 is what lets a prime be left out entirely.
How many factors does 360 have, and what do they add up to?
360 = 2³ × 3² × 5
Count = (3+1)(2+1)(1+1) = 4 × 3 × 2 = 24
Sum = (1+2+4+8)(1+3+9)(1+5) = 15 × 13 × 6 = 1170
Note the sum needs each bracket multiplied, not added. Adding gives 34, which is nonsense — the sum of factors must be bigger than N itself.
Special families of factors — all from the same factorisation
Remember 0 counts as even and as a multiple of 3 — that is why 1 is always both a perfect square factor and a perfect cube factor.
How many factors of 4500 are perfect squares?
4500 = 45 × 100 = (3²×5) × (2²×5²) = 2² × 3² × 5³
Even exponents available:
for 2 → 0 or 2 → 2 choices
for 3 → 0 or 2 → 2 choices
for 5 → 0 or 2 (3 is odd, so the third 5 is wasted) → 2 choices
Total = 2 × 2 × 2 = 8: they are 1, 4, 9, 25, 36, 100, 225, 900.
How many factors of 28 × 34 are multiples of 24 but not of 48?
A factor looks like 2x3y.
Multiple of 24 = 2³×3 → needs x ≥ 3 and y ≥ 1.
Not a multiple of 48 = 2⁴×3 → since y ≥ 1 already, the only escape is x < 4.
Both conditions land on the same prime: x ≥ 3 and x < 4 → x = 3 exactly. y is free: 1, 2, 3 or 4 → 4 choices.
Answer = 1 × 4 = 4 (they are 24, 72, 216, 648).
The idea to keep: when two conditions touch the same prime, merge them into one range before counting. Never count them separately and multiply.
Reading a number from its factor count — used in "how many numbers have exactly k factors" questions
Method: write the target count as a product, subtract 1 from each part to get the exponents, then hunt inside the given range.
An odd number of factors means the number is a perfect square — because factors pair up (d with N/d) and only a square has a middle factor pairing with itself. 36 has 9 factors; 105² = 11025 has 27. Use this as a free check on any factor-count answer.
HCF (highest common factor) = the biggest number that divides both.
LCM (lowest common multiple) = the smallest number that both divide into.
From prime factorisation: HCF takes the lowest power of each shared prime; LCM takes the highest power of every prime that appears.
24 = 2³×3, 36 = 2²×3² → HCF = 2²×3 = 12, LCM = 2³×3² = 72.
The product rule: HCF × LCM = first number × second number.
It works because for each prime, the HCF takes the smaller power and the LCM the larger — between them they use both original powers exactly once.
Warning: this rule holds for two numbers only, never for three.
The product of two numbers is 4032 and their HCF is 12. Find the LCM.
LCM = 4032 ÷ 12 = 336.
Check it is possible: the HCF must divide the LCM, and 336 ÷ 12 = 28 ✔ (An actual pair is 48 and 84.)
When numbers are given as a ratio. If two numbers are in the ratio m : n in lowest terms, write them as mk and nk. Then:
HCF = k and LCM = m × n × k
Ratio 7 : 9 with HCF 6 → numbers are 42 and 54, LCM = 7×9×6 = 378.
Ratio 4 : 5 with LCM 120 → 20k = 120 → k = 6 = HCF.
This only works if the ratio is already reduced. For 8 : 10 the terms share a 2 and k is not the HCF.
For co-prime numbers, LCM = the product (since HCF = 1). Two co-prime numbers with LCM 143: since 143 = 11 × 13, they are 11 and 13.
Least three-digit number leaving remainder 5 when divided by 24, 36 and 54.
If N leaves 5 with all three, then N − 5 is divisible by all three, so N − 5 is a multiple of the LCM.
24 = 2³×3, 36 = 2²×3², 54 = 2×3³ → LCM = 2³×3³ = 216.
Family: N = 216k + 5 → 5, 221, 437, …
First three-digit member = 221.
Greatest number that divides 43, 91 and 183 leaving the same remainder each time.
The remainder is unknown, so subtract in pairs — the unknown cancels:
91 − 43 = 48, 183 − 91 = 92, 183 − 43 = 140.
Answer = HCF(48, 92, 140) = 4. (Check: each leaves remainder 3.)
In problem type 1 above, r itself is technically an answer — 5 divided by 24 gives quotient 0 and remainder 5. That is why good questions ask for the least three-digit such number. If a question does not say it, and r is one of the options, the question is faulty; pick the LCM-based answer, which is what is intended.
Division algorithm: Dividend = Divisor × Quotient + Remainder, and the remainder is always smaller than the divisor.
A number leaving remainder r on division by d looks like N = dk + r. Adding d changes the quotient but never the remainder, so all such numbers sit d apart: r, r+d, r+2d, …
Remainders behave like numbers. You may replace any number by its remainder before adding, subtracting or multiplying.
Written as: if a leaves remainder x and b leaves remainder y on division by d, then a×b leaves the same remainder as x×y, and a+b the same as x+y. (Reduce again at the end if the result crosses d.)
A number leaves remainder 3 when divided by 7. What does its square leave?
Just square the remainder: 3² = 9, and 9 = 7 + 2, so the remainder is 2.
Proof if you want it: N = 7k+3, so N² = 49k² + 42k + 9. The first two terms are multiples of 7; only the 9 is left, and 9 leaves 2.
The negative remainder trick — the biggest time-saver here.
If a number is 1 less than the divisor, treat it as −1.
Then powers become trivial: (−1)even = +1 and (−1)odd = −1.
Remainder when 121 + 122 + … + 1210 is divided by 13.
12 ≡ −1 (mod 13), so the terms are −1, +1, −1, +1, … alternating.
Pair them: (−1 + 1) + (−1 + 1) + … Each pair gives 0.
There are 10 terms = 5 complete pairs, nothing left over → remainder 0.
Same working handles 6+6²+… ÷ 7, 8+8²+… ÷ 9, 10+10²+… ÷ 11, 14+14²+… ÷ 15.
If the number of terms were odd, one −1 would be left unpaired and the answer would be d − 1. Always check whether the count is even.
How many numbers from 1 to 400 leave remainder 5 with 9 and remainder 4 with 7?
Hunt the first one: numbers leaving 5 with 9 are 5, 14, 23, 32 … and 32 also leaves 4 with 7 ✔
Because 9 and 7 are co-prime, the pattern then repeats every LCM(9,7) = 63.
Family: 32, 95, 158, 221, 284, 347 (next would be 410, too big) → 6 numbers.
When counting "how many numbers from 1 to X leave remainder r on division by d", the smallest member is r itself — because r ÷ d has quotient 0 and remainder r. Dropping it is the commonest way to lose one from the count. And count with k = 0 to the last k, so the answer is (last k) + 1.
The single most-asked idea in this chapter. It takes ten minutes to learn and never leaves you.
Only the last digit of the base matters. As you take higher powers, the last digit repeats in a fixed block called the cycle.
| Last digit of base | Powers 1, 2, 3, 4 end in… | Cycle length |
|---|---|---|
| 0, 1, 5, 6 | same digit every time | 1 |
| 4 | 4, 6, 4, 6 | 2 |
| 9 | 9, 1, 9, 1 | 2 |
| 2 | 2, 4, 8, 6 | 4 |
| 3 | 3, 9, 7, 1 | 4 |
| 7 | 7, 9, 3, 1 | 4 |
| 8 | 8, 4, 2, 6 | 4 |
The method, in three steps
Remainder 0 means the block finished exactly, so you land on the end of the cycle — not the beginning. This one point causes more wrong answers than anything else in the section.
Dividing the power by 4 the fast way: only the last two digits of the power matter, since 100 is a multiple of 4. For 7289, just do 89 ÷ 4 → remainder 1.
Odd/even is enough for bases ending in 4 and 9: odd power → 4 or 9; even power → 6 or 1.
Nothing to do for bases ending in 0, 1, 5 or 6 — the digit never changes.
Unit digit of 2100 + 3100.
2 has cycle 2, 4, 8, 6. 100 ÷ 4 gives remainder 0 → take the last entry → 6.
3 has cycle 3, 9, 7, 1. Remainder 0 again → last entry → 1.
6 + 1 = 7 → unit digit 7.
If you had read remainder 0 as "first entry" you would get 2 + 3 = 5 — and 5 will be sitting there as an option.
Unit digit of (1717)17.
First clear the bracket: powers of a power multiply → 17289.
Base ends in 7 → cycle 7, 9, 3, 1. Power 289 → 89 ÷ 4 leaves 1 → first entry → 7.
Reading it as 1734 (adding instead of multiplying) would give 9. Multiply the exponents.
Products and sums
n! ("n factorial") = 1 × 2 × 3 × … × n. So 5! = 120.
A zero at the end of a number comes from a factor of 10, and 10 = 2 × 5. So the number of trailing zeros = the number of (2 × 5) pairs inside.
In any factorial the 2s are far more plentiful than the 5s — every second number gives a 2, but only every fifth gives a 5. So just count the 5s.
The counting formula (Legendre's). Number of 5s in n! =
⌊n/5⌋ + ⌊n/25⌋ + ⌊n/125⌋ + ⌊n/625⌋ + …
The brackets ⌊ ⌋ mean "drop the decimal part". Keep going until the divisor is bigger than n.
Why more than one term? 25 carries two 5s, 125 carries three, 625 carries four — the first term counts each of them only once, so the later terms pick up the extras.
Zeros at the end of 165!
⌊165/5⌋ = 33
⌊165/25⌋ = 6
⌊165/125⌋ = 1
625 > 165, stop.
Total = 33 + 6 + 1 = 40 zeros.
Zeros at the end of 1000!
200 + 40 + 8 + 1 = 249. Worth remembering as a landmark.
Notice the fourth term: 1000 is past 625 = 54, so a fourth term exists. Stopping at three gives 248 — which will be an option.
How many terms will you need? Compare n with the powers of 5: 5, 25, 125, 625, 3125.
Glance at n first and you will never stop one term early.
Product of even numbers. 2×4×6×…×2n = 2n × n!, because you can pull a 2 out of every term. The 2s are then obvious and you count the 5s inside n! as usual.
Example: zeros at the end of the product of the first 25 even numbers = zeros of 225 × 25! → 5s in 25! = 5 + 1 = 6.
For the highest power of a composite like 6 or 12, split it into primes and take the weaker one. Highest power of 6 in 20!: count 3s (6+2 = 8) and count 2s (10+5+2+1 = 18). Since 6 = 2×3, each 6 needs one of each, so the answer is the smaller — 8.
| Squares 11–30 | 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441, 484, 529, 576, 625, 676, 729, 784, 841, 900 |
|---|---|
| Cubes 1–20 | 1, 8, 27, 64, 125, 216, 343, 512, 729, 1000, 1331, 1728, 2197, 2744, 3375, 4096, 4913, 5832, 6859, 8000 |
| Powers of 2 | 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096 … 215 = 32768, 220 = 1048576 |
A perfect square never ends in 2, 3, 7 or 8. And if a square ends in zeros, it ends in an even number of them — so 8000 (three zeros) cannot be a perfect square. One glance rules out options.
Least number to add or subtract to reach a perfect square.
Least number subtracted from 8000 to make a perfect square.
√8000 ≈ 89.4. Round down: 89² = 7921.
8000 − 7921 = 79.
(If the question had said "added", we would use 90² = 8100 and the answer would be 100.)
Least number added to 5000 to make a perfect cube.
17³ = 4913 and 18³ = 5832. We need the cube above, so 5832 − 5000 = 832.
Square roots of decimals. A square root halves the number of decimal places.
Trick to see it: write as a fraction. 0.0625 = 625/10000, and √625 / √10000 = 25/100 = 0.25.
Note: the root of a decimal below 1 is larger than the decimal itself. √0.09 = 0.3, not 0.045.
Counting the digits of a big number.
A surd is a root that does not come out exact — √2, √5, √300.
Values worth memorising: √2 ≈ 1.414, √3 ≈ 1.732, √5 ≈ 2.236, √6 ≈ 2.449, √7 ≈ 2.646.
Conjugates. √a + √b and √a − √b are conjugates. Multiply them and the surds vanish:
(√a + √b)(√a − √b) = a − b
This is the difference-of-squares identity, and it is why conjugates clear surds out of a denominator.
Simplify 2 / (√5 − √3).
Multiply top and bottom by the conjugate √5 + √3:
denominator becomes 5 − 3 = 2, numerator becomes 2(√5 + √3).
The 2s cancel → √5 + √3.
When x is a conjugate fraction. If x = (√a + √b) / (√a − √b), then 1/x is the same fraction flipped over. So:
For x = 1/(√6 − √5): rationalising gives x = √6 + √5 and 1/x = √6 − √5. So x − 1/x = 2√5 and x + 1/x = 2√6.
Placing a surd between two integers. Find the perfect squares on either side.
√300: 17² = 289 and 18² = 324, and 289 < 300 < 324, so the value is between 17 and 18.
Or simplify first: √300 = √(100×3) = 10√3 ≈ 17.32.
Comparing roots with different indices — raise them all to the LCM of the index numbers.
Compare √3, ∛5, 6√20: indices 2, 3, 6 → LCM 6. Raise each to the 6th power: 3³ = 27, 5² = 25, 20. Since 27 is biggest, √3 is the largest.
Note the big number under a root can still be the smallest — a higher index pulls the value down harder than a bigger number pushes it up.
Roots split over multiplication and division but never over addition or subtraction:
√(a×b) = √a × √b ✔ √(a+b) ≠ √a + √b ✘
So √1.69 − √0.09 = 1.3 − 0.3 = 1.0, not √1.60.
The four sums to memorise
Sum of an evenly spaced list (AP):
Sum = (number of terms / 2) × (first + last) and number of terms = (last − first)/gap + 1
Or think of it as average × count, where the average of an evenly spaced list is (first + last)/2.
Find 7 + 14 + 21 + … + 140.
Pull the 7 out: 7(1 + 2 + … + 20) = 7 × (20×21/2) = 7 × 210 = 1470.
Or by the AP formula: 20 terms, (20/2)(7 + 140) = 10 × 147 = 1470.
The usual mistake: taking the number of terms as 140. The count is 140 ÷ 7 = 20.
Consecutive numbers — centre on the middle.
If the count is odd, the average is the middle number. Five consecutive multiples of 9 summing to 225 → middle = 225/5 = 45, so the run is 27, 36, 45, 54, 63 and the largest is 63.
Name the numbers as m−2d, m−d, m, m+d, m+2d and everything except the middles cancels — that is why this works.
If the count is even there is no middle term; the average falls between the two central ones. Six consecutive multiples of 4 summing to 156 → average 26, which sits between 24 and 28, so the run is 16 … 36.
Telescoping sums. When each term is 1/(k×(k+1)), split it:
1/(k(k+1)) = 1/k − 1/(k+1)
Then almost everything cancels:
1/(1×2) + 1/(2×3) + … + 1/(9×10) = (1 − ½) + (½ − ⅓) + … + (1/9 − 1/10) = 1 − 1/10 = 9/10
General result: the sum up to 1/(n(n+1)) is n/(n+1).
Sum of k consecutive integers = kn + k(k−1)/2. So:
Multiples in a range: the count of multiples of d from 1 to N is ⌊N/d⌋ — divide and drop the decimal.
Multiples of 8 up to 200 → ⌊200/8⌋ = 25.
Two conditions at once (inclusion–exclusion).
divisible by a OR b = ⌊N/a⌋ + ⌊N/b⌋ − ⌊N/LCM⌋
The overlap is subtracted once because those numbers were counted twice.
Exactly one of the two (the usual phrasing is "by a or b but not by their LCM"):
⌊N/a⌋ + ⌊N/b⌋ − 2×⌊N/LCM⌋
Subtract the overlap twice — once to stop double counting, once more to throw those numbers out altogether.
From 1 to 990, how many are divisible by 9 or 11 but not by 99?
⌊990/9⌋ = 110, ⌊990/11⌋ = 90. Since 9 and 11 are co-prime, LCM = 99, and ⌊990/99⌋ = 10.
Exactly one = 110 + 90 − 20 = 180.
Check: 9 only = 100, 11 only = 80, and 100 + 80 = 180 ✔
The tempting wrong answer is 190 — that is the "or" count with the final exclusion forgotten.
The overlap uses the LCM, not the product. For 6 and 10 the overlap is 30, not 60 — they share a 2. Only for co-prime pairs is the LCM equal to the product.
Fixed digit in a fixed place. Count choices for each place and multiply. The leading digit cannot be 0 — and that rule bites exactly once.
How many 3-digit numbers have a digit sum of 7?
Write it as a + b + c = 7 with a ≥ 1. Substitute a′ = a − 1 so all three can be 0 or more: a′ + b + c = 6.
Distributing 6 identical units among 3 places gives C(6+2, 2) = C(8,2) = 28.
Slower but safer check: with a = 1 there are 7 ways, a = 2 gives 6, then 5, 4, 3, 2, 1 → 7+6+5+4+3+2+1 = 28 ✔
Complementary counting — count the opposite when the direct count splits into messy cases.
"In how many 3-digit numbers is the product of the digits zero?" means at least one digit is 0. Count instead the ones with no zero: 9×9×9 = 729. Total 3-digit numbers = 900. Answer = 900 − 729 = 171.
Binary uses only 0 and 1. The place values, read from the right, are 1, 2, 4, 8, 16, 32, 64, 128, … — each double the last (instead of 1, 10, 100 in our normal decimal system).
Binary to decimal: 1101101
Write the place values above the digits and add the ones that are switched on:
64 + 32 + 8 + 4 + 1 = 109
Decimal to binary: 96
Divide by 2 repeatedly and read the remainders bottom to top:
96→48 r0, 48→24 r0, 24→12 r0, 12→6 r0, 6→3 r0, 3→1 r1, 1→0 r1 → 1100000
Or by subtraction: 96 = 64 + 32, so switch on those two places.
Facts that answer whole questions instantly
For arithmetic, convert to decimal, calculate, convert back. It is faster and far safer than borrowing and carrying in binary, where a carry is worth 2 and not 10.
10000 − 111 → 16 − 7 = 9 → 1001
Then check by parity: 16 even minus 7 odd must be odd, and 1001 ends in 1 ✔
| Identity | What it is for |
|---|---|
| a² − b² = (a+b)(a−b) | Difference of squares; conjugate surds; consecutive-multiple differences |
| (a+b)² = a² + 2ab + b² | Links sum, product and sum-of-squares |
| (a−b)² = a² − 2ab + b² | Same, for a difference |
| (a+b)² − (a−b)² = 4ab | Recovers the product from sum and difference |
| (a+b)² = (a−b)² + 4ab | Finds a+b when you know a−b and ab |
| a² + b² = ((a+b)² + (a−b)²) / 2 | Sum of squares from sum and difference |
If a − b = 4 and ab = 45 (both positive), find a + b.
(a+b)² = (a−b)² + 4ab = 16 + 180 = 196 → a + b = 14.
(Then a = 9, b = 5. Check: 9−5 = 4 ✔, 9×5 = 45 ✔)
The x + 1/x ladder. Squaring introduces a +2 every time, because x × 1/x = 1.
Note the sign: with a plus given you subtract 2; with a minus given you add 2.
If a + 1/a = 4, find a⁴ + 1/a⁴.
Step up twice:
a² + 1/a² = 4² − 2 = 14
a⁴ + 1/a⁴ = 14² − 2 = 194
Answering 196 means you forgot the last −2; answering 254 means you squared 16 instead of 14.
Two-digit numbers and their reverse. Write the number as 10a + b. Then:
"Reversing decreases the number by 36" → 9(a−b) = 36 → a − b = 4. With a + b = 10 you get a = 7, b = 3, so the number is 73.
Watch the direction: 37 has the same digit sum and the same size of change, but there the reverse is bigger. That is the intended trap.
These are number-system word problems, not percentage or profit sums. Each has a fixed shape — learn the shape and the numbers become easy.
1. Two kinds of coins or notes. Count equation + value equation.
45 coins of one rupee and two rupees, worth Rs 68 in all. How many two-rupee coins?
a + b = 45 and a + 2b = 68. Subtract → b = 23.
Faster: if all 45 were one-rupee coins the value would be 45; the extra 68 − 45 = 23 comes from the two-rupee coins, each adding 1 extra.
2. Marks with a penalty. "Attempts all" means correct + wrong = total, with no third group.
4 marks per correct, 1 deducted per wrong, all 50 attempted, scored 120.
4c − (50 − c) = 120 → 5c = 170 → c = 34.
Faster: all correct = 200. Each mistake costs 4 + 1 = 5 (the 4 not earned plus the 1 deducted). Shortfall 80 → 80/5 = 16 wrong → 34 correct.
wrong = (maximum − score) / (marks per right + penalty)
3. Ages. The golden rule: after t years everybody ages by t. Never age one person alone.
Father is 7 times the son now; in 6 years he will be 4 times.
7s + 6 = 4(s + 6) → 3s = 18 → son is 6, father 42.
Check that always works: the age gap never changes. 42 − 6 = 36 now, 48 − 12 = 36 later ✔
Formula: if the father is m times now and n times after t years, son's age = t(n−1)/(m−n).
4. Heads and legs. Hens 2 legs, cows 4; cycles 2 wheels, rickshaws 3.
36 heads, 100 legs, hens and cows. → h + c = 36, 2h + 4c = 100 → h = 22.
Faster: if all were hens, legs = 72. Surplus 28, and each cow adds 2 extra → 14 cows, 22 hens.
5. Sum-and-relation. Two numbers, a sum and a comparison.
Sum 108; the larger is 8 more than 3 times the smaller.
Write the larger two ways: 108 − s and 3s + 8. Set them equal → 4s = 100 → s = 25, larger = 83.
Read carefully: "8 more than 3 times the smaller" is 3s + 8, not 3(s + 8).
6. Fractional parts of one number. "5 out of every 8 parts" simply means five-eighths.
5/8 of a number exceeds 3/5 of it by 6.
Subtract the fractions first: 5/8 − 3/5 = 25/40 − 24/40 = 1/40.
So (1/40)x = 6 → x = 240.
number = given difference ÷ difference of the fractions
7. Ratios. Turn m : n into mx and nx — the ratio fixes proportions, x fixes size.
Two-digit numbers with a digit-sum condition. Write 10a + b and simplify.
When a family survives the first condition, the second condition (digits differ by 1, digits add to 10, and so on) picks the single answer.
Most lost marks in this chapter are not calculation errors. They are reading errors. These six come up again and again.
Which number does it want? "The larger", "the middle one", "the total", "how many more". You solve for the smaller and tick it out of habit. Underline the final phrase before you start.
Face value or place value? Sum of digits, sum of place values, difference of place values — three different answers from the same number, and all three will be options.
Remainder 0 in cyclicity. A power divisible by the cycle length lands on the last entry, not the first. 2100 ends in 6, not 2.
"Always divisible by" with two true options. If both 6 and 12 always work, the question means the largest. Test the smallest value of n first — it usually pins the answer exactly.
Negative possibilities. "Two numbers differ by 3 and their squares add to 117" is satisfied by 9 and 6 — and also by −9 and −6. If the options contain only positives, the question means positives. Check the options before worrying.
Two options that are the same value. A ratio may appear reduced in one option and unreduced in another — 8 : 15 and 56 : 105 are the same thing. Always reduce your answer before matching it to the list, and pick the reduced form.
Three free checks to run on any answer
The night before the exam, read only this page.
| Topic | Formula |
|---|---|
| Number of factors | (a+1)(b+1)(c+1)… from N = paqbrc |
| Sum of factors | (1+p+…+pa)(1+q+…+qb)… |
| Product of factors | N(number of factors)/2 |
| Square factors | (⌊a/2⌋+1)(⌊b/2⌋+1)… | Cube factors: ⌊a/3⌋+1 each |
| HCF × LCM | = product of the two numbers (two numbers only) |
| Ratio m : n with HCF k | numbers mk, nk | LCM = mnk |
| Same remainder r with a, b, c | N = LCM(a,b,c) × k + r |
| Same unknown remainder | HCF of the differences |
| Trailing zeros of n! | ⌊n/5⌋ + ⌊n/25⌋ + ⌊n/125⌋ + ⌊n/625⌋ |
| 1 + 2 + … + n | n(n+1)/2 | odds: n² | evens: n(n+1) |
| Sum of squares | n(n+1)(2n+1)/6 |
| AP sum | (count/2) × (first + last); count = (last − first)/gap + 1 |
| Telescoping | Σ 1/(k(k+1)) up to n = n/(n+1) |
| Exactly one of a, b | ⌊N/a⌋ + ⌊N/b⌋ − 2⌊N/LCM⌋ |
| Two-digit reversal | difference = 9(a−b); sum = 11(a+b) |
| x² + 1/x² | (x + 1/x)² − 2 | (x − 1/x)² + 2 |
| a + b from a − b and ab | (a+b)² = (a−b)² + 4ab |
| Conjugate surds | (√a + √b)(√a − √b) = a − b |
| Marks with penalty | wrong = (max − score) / (marks per right + penalty) |
If you can do these eight without notes, you are ready for the sets. Answers below.
Answers: 1) 18 2) 3 3) 33 4) 428 5) 275 6) 4 7) 56 8) 11
Number System is the chapter that deals with numbers and their properties — types of numbers, divisibility, factors, HCF and LCM, remainders, unit digits and related ideas. It is the base for several other quant chapters, so it is usually studied first.
Yes. Questions from Number System appear in every SSC CGL Tier-1 paper, and the concepts also feed into simplification, LCM-HCF word problems and data-based questions. It is high scoring because most questions need one rule and very little calculation.
Start with types of numbers and prime factorisation. Prime factorisation alone unlocks factors, divisor counting, HCF, LCM and perfect-square questions — roughly a third of the chapter.
Use the cyclicity table. Take the last digit of the base, divide the power by the cycle length (usually 4), and read the cycle at that remainder. A remainder of 0 means the last entry of the cycle, not the first.
Write the number as pa × qb × rc and multiply (a+1)(b+1)(c+1). For 360 = 2³ × 3² × 5 the count is 4 × 3 × 2 = 24.
After finishing a section here, attempt the matching Number System practice test on TrickySSC. The tests are graded across three levels, and every question carries a step-by-step solution.
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