Chapter 1 of 19

SSC CGL Quantitative Aptitude★ Tier 1: 1–2 Q (2–4 marks) · Tier 2: 2–3 Q (6–9 marks)

Number System

1–2 questions in Tier 1 and 2–3 in Tier 2, and the mental machinery behind half the quant paper. Divisibility rules, factor counting, HCF–LCM identities and remainder cycles — four toolkits that turn "big ugly number" questions into 30-second wins.

⏱ 22 min readModerate difficulty✓ Free · exam-pattern aligned

Timed practiceMCQs from this chapter with instant scoring

FlashcardsFormulas & traps on spaced repetition

SSC CGL PYQsSolved previous-year questions

By the end of this chapter you'll be able to…

1Apply the complete SSC divisibility rule set, including composite divisors split into coprime factors (72 = 8 × 9, 88 = 8 × 11)
2Count and sum the factors of any number from its prime factorisation, and use the odd-factor-count ⇔ perfect-square signal
3Use HCF × LCM = product, and match word problems to the four HCF/LCM templates (toll together, greatest divisor, same remainder, least number with remainder)
4Compute remainders of large powers using cycles, negative remainders, and the exponent-reduction shortcut
5Find unit digits via mod-10 cycles of length 1, 2, or 4

💡

Why this chapter matters in SSC CGL

Number system is double-counted in your score: it contributes its own 1–3 questions per tier, and its toolkits (divisibility, LCM instincts, remainder cycles) speed up simplification, ratio, averages, and time–work questions across the paper. Candidates who compute divisibility by actually dividing lose 30–60 seconds per question — across a 15-minute Tier-1 quant section, that is the difference between attempting 20 questions and attempting 25.

Number System — SSC CGL Quantitative Aptitude

Number system questions look scary — "remainder when $3^{100}$ is divided by 13" — but every one of them runs on a small set of mechanical toolkits: divisibility rules, the factor-counting formula, the HCF×LCM identity, and remainder cycles. Learn the four toolkits and the scary numbers become routine.

1. What SSC actually asks

Tier 1: 1–2 Q · Tier 2: 2–3 Q. The recurring types, in rough order of frequency:

Divisibility — find a missing digit so a number is divisible by 3, 8, 9, 11, or a composite like 72 or 88.
Remainders — big powers mod small numbers, unit digits, "a number leaves remainder $r$ on division by $d$ …".
HCF & LCM — identities, bells-tolling-together, greatest-number-that-divides word problems.
Factors — count, sum, or parity of factors of a given number.

Beyond its own questions, this chapter is infrastructure: simplification, ratio, averages and time–work all lean on fast divisibility and LCM instincts.

2. Divisibility rules — the complete SSC set

Divisor	Rule
2, 5, 10	Last digit
4, 25	Last two digits
8, 125	Last three digits
3, 9	Digit sum divisible by 3 / 9
11	(Sum of digits at odd places) − (sum at even places) divisible by 11 (0 counts)
6	Rules for 2 and 3
12	Rules for 3 and 4
72	Rules for 8 and 9
88	Rules for 8 and 11
7, 13	Rarely tested directly; use actual division or the 1001 trick ( $7 \times 11 \times 13 = 1001$ )

The composite trick: for a composite divisor, split into coprime factors and apply both rules. $72 = 8 \times 9$ works; $72 = 6 \times 12$ does not (6 and 12 share a factor).

3. Factors: count, sum, and the perfect-square signal

Write $N = a^{p} b^{q} c^{r}$ (prime factorisation). Then:

Number of factors $= (p + 1) (q + 1) (r + 1)$
Sum of factors $= \frac{a ^{p + 1} - 1}{a - 1} \cdot \frac{b ^{q + 1} - 1}{b - 1} \cdot \frac{c ^{r + 1} - 1}{c - 1}$
Odd number of factors ⇔ $N$ is a perfect square (factors pair up except $N$ )

Example. $720 = 2^{4} \cdot 3^{2} \cdot 5^{1}$ → factors $= 5 \times 3 \times 2 = 30$ .

4. HCF and LCM — identities and word-problem templates

$HCF \times LCM = product of the two numbers (two numbers only!)$

For fractions: $HCF = \frac{HCF of numerators}{LCM of denominators}$ , $LCM = \frac{LCM of numerators}{HCF of denominators}$ .

The four word-problem templates:

Question phrase	Compute
"Bells toll together / lights flash together"	LCM of intervals
"Greatest number that divides $a, b, c$ exactly"	HCF $(a, b, c)$
"Greatest number that divides $a, b, c$ leaving the same remainder"	HCF of the differences
"Least number which when divided by $a, b, c$ leaves remainder $r$ in each case"	LCM $(a, b, c) + r$

5. Remainders — the workhorse toolkit

Rule 1 (break it down): remainders multiply and add. To find $N mod d$ , replace every big piece by its remainder as you go.

Rule 2 (find the cycle): powers repeat mod $d$ . Locate the power where the remainder returns to 1, then reduce the exponent.

$2^{47} mod 7 : 2^{3} = 8 \equiv 1 \Rightarrow 2^{47} = (2^{3})^{15} \cdot 2^{2} \equiv 1 \cdot 4 = 4$

Rule 3 (negative remainders speed things up): $d - 1 \equiv - 1$ . So $2 5^{81} mod 26 \equiv (- 1)^{81} = - 1 \equiv 25$ .

Rule 4 (unit digits are just mod 10): cycles have length at most 4 —

Unit digit of base	Cycle
0, 1, 5, 6	constant
4, 9	length 2 (4,6 / 9,1)
2, 3, 7, 8	length 4

Reduce the exponent mod 4 (use 4, not 0, when it divides exactly).

Rule 5 (divisor factors): if $N mod 56 = 29$ , then $N mod 8$ : since $8 ∣ 56$ , answer is $29 mod 8 = 5$ . Works only when the second divisor divides the first.

6. Solved PYQ-style examples

Q1. Largest 4-digit number divisible by 88? Solution. $9999 \div 88 = 113.6 \dots$ → $88 \times 113 = 9944$ . 9944.

Q2. Unit digit of $7^{105}$ ? Solution. 7-cycle: 7, 9, 3, 1 (length 4). $105 mod 4 = 1$ → first entry → 7.

Q3. The greatest number dividing 62, 132 and 237 leaving the same remainder? Solution. Differences: $132 - 62 = 70$ , $237 - 132 = 105$ , $237 - 62 = 175$ . HCF $(70, 105, 175) =$ 35.

Q4. HCF of two numbers is 12, LCM is 336, one number is 84. Other? Solution. $\frac{12 \times 336}{84} = 48$ . 48. (Sanity check: HCF(84, 48) = 12 ✓.)

Q5. $3^{100} mod 13$ ? Solution. $3^{3} = 27 \equiv 1 (mod 13)$ . $100 = 3 \times 33 + 1$ → $3^{100} \equiv 3^{1} = 3$ . 3.

7. Exam protocol

Digit-sum first: it settles divisibility by 3 and 9 instantly and catches arithmetic slips in your own work.
For missing-digit divisibility, write the two coprime conditions as equations before touching options.
For power-remainder questions, hunt for the exponent where the remainder becomes 1 (or −1) — it is almost always ≤ 6 steps away.
Two numbers with given HCF? Write them as $H x, H y$ with $x, y$ coprime — most "sum/product with HCF" questions fall in one line.
Word problem? Match it to one of the four HCF/LCM templates before computing anything.

Key formulas & results

Everything to memorise for the exam hall, in one card. Screenshot this for revision.

Number of factors

N = a^{p} b^{q} c^{r} \Rightarrow factors = (p + 1) (q + 1) (r + 1)

Prime factorise first. Odd factor count means N is a perfect square.

Sum of factors

σ (N) = \frac{a ^{p + 1} - 1}{a - 1} \cdot \frac{b ^{q + 1} - 1}{b - 1} \dots

One geometric-series bracket per prime.

HCF–LCM identity

HCF (a, b) \times LCM (a, b) = a \times b

Two numbers only — fails for three or more.

Same-remainder divisor

Greatest divisor of a, b, c (same remainder) = HCF (b - a, c - b, c - a)

Differences cancel the common remainder.

Negative remainder shortcut

(d - 1)^{n} mod d = {1 d - 1 n even n odd

Treat d−1 as −1. Kills questions like 25^81 mod 26 in five seconds.

Divisibility by 11

11 ∣ N ⟺ (odd-place digit sum) - (even-place digit sum) \equiv 0 (mod 11)

Count places from the right; a difference of 0 also qualifies.

SSC CGL test series

Ready to be tested on this?

Number System shows up in every SSC CGL paper — sit a timed one and find out where you actually stand.

1–2Qs in Tier 1

Mediumpriority

Full SSC CGL mocks, real clock
Quantitative Aptitude-only sectionals & chapter tests
All-India rank + topic-level analysis

Free forever · solver-verified questions

⚠️

Traps SSC CGL sets — and how to dodge them

These are the exact option-traps and misreads that cost marks under negative marking.

WATCH OUT

✗ Splitting a composite divisor into non-coprime factors — testing 72 as 6 × 12.

✓ The split must be coprime: 72 = 8 × 9. A number like 24 passes both the 6-rule and the 12-rule but is not divisible by 72.

WATCH OUT

✗ Using HCF × LCM = product for three numbers.

✓ The identity holds for exactly two numbers. For three, work from prime factorisations directly.

WATCH OUT

✗ Reducing the exponent mod 4 for unit digits and using cycle position 0.

✓ When the exponent is an exact multiple of the cycle length, use the LAST entry of the cycle, not the first. 2^4 ends in 6, not 2.

WATCH OUT

✗ In 'same remainder' problems, taking HCF of the numbers instead of HCF of their differences.

✓ The common remainder survives in each number but cancels in the differences — HCF of differences is the answer.

WATCH OUT

✗ Assuming N mod d determines N mod e for any smaller e.

✓ It works only when e divides d. From N mod 56 = 29 you can get N mod 8 (= 5), but not N mod 6.

WATCH OUT

✗ Forgetting that 0 is divisible by everything when checking the 11-rule.

✓ An alternating difference of 0 means divisible by 11. Rejecting it costs the question.

Exam-pattern practice

PYQ-style questions with full solutions. Work through them as a readiness check — mark yourself honestly and get your gap report at the end.

Readiness check

Are you exam-ready for Number System?

10 problems from this chapter. Try each one, reveal the worked solution, mark yourself honestly — get your gap report at the end.

10 questions~7 min

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•Composite divisibility needs a COPRIME split: 72 = 8 × 9, 88 = 8 × 11 — never 6 × 12
•11-rule: (odd places) − (even places) from the right, multiple of 11 (including 0)
•N = aᵖbᵠcʳ → factor count (p+1)(q+1)(r+1); odd count ⇔ perfect square
•HCF × LCM = product — for exactly two numbers
•Same remainder on a, b, c → answer is HCF of the differences
•Least number leaving remainder r with each of a, b, c → LCM(a,b,c) + r
•Power remainders: find where the cycle hits 1 (or −1), then reduce the exponent
•d − 1 ≡ −1 (mod d): (d−1)^odd leaves d−1, (d−1)^even leaves 1
•Unit-digit cycles: 0/1/5/6 constant, 4/9 length 2, 2/3/7/8 length 4 — exact multiples use the cycle's LAST entry
•N mod d gives N mod e only when e divides d

SSC CGL question blueprint

How this topic is asked, tier by tier — so you can prep to the pattern.

Typical weightage: 13

Question style	Marks each	Typical count	What it tests
Tier 1 — divisibility / unit digit / HCF–LCM	2–4 (1–2 Q × 2 marks)
Tier 2 — remainders, factor counting, word templates	6–9 (2–3 Q × 3 marks)

Prep strategy

Memorise the divisibility table and the four HCF/LCM word templates cold — recall, not derivation
Drill 20 power-remainder questions until cycle-hunting is reflexive
Practise missing-digit questions with composite divisors (72, 88, 132) using coprime splits
Weekly timed set: 15 mixed number-system questions in 12 minutes

Exam-hall strategy

Battle-tested tips from mentors and toppers for this topic under the sectional clock.

Digit-sum every candidate number first — it resolves divisibility by 3 and 9 instantly and doubles as a check on your own arithmetic.
For missing-digit problems, write both coprime-factor conditions as small equations before glancing at the options.
In power-remainder questions, spend the first ten seconds hunting the cycle: multiply the base until the remainder is 1 or −1.
For HCF word problems, write the numbers as Hx and Hy with x, y coprime — sums and products then factor cleanly.
If a question resists both toolkits for 60 seconds, mark and move — number system has no partial credit for elegant struggle.

Beyond the exam

Where this skill shows up in the job you're competing for — and in life.

Cryptography

RSA encryption is modular exponentiation at industrial scale — the same remainder-cycle arithmetic you use on 3¹⁰⁰ mod 13, run on 2048-bit numbers.

Scheduling and synchronisation

When do two cron jobs with different periods collide? When do traffic lights sync? Every such question is the bells-tolling LCM template.

Checksums and error detection

The 11-divisibility rule is the exact mechanism behind ISBN check digits; digit sums power the checksums in card numbers and barcodes.

Where else this topic is tested

Prepare once, score in every exam that asks it.

SSC CHSL2–3 Q — same templates, gentler numbers

SSC CPO2–3 Q — heavy on divisibility and LCM word problems

RRB NTPC3–4 Q — unit digits and HCF/LCM staples

IBPS Clerk1–2 Q — usually inside simplification sets

📝 My notes

Saved on this device — sign in to sync · how this works

0/600

Questions aspirants ask

Pulled from the Q&A community and mentor sessions.

Usually 1–2 in Tier 1 and 2–3 in Tier 2 Paper 1, drawn from divisibility, remainders, HCF/LCM and factor counting. Its real value is larger: the same toolkits speed up simplification, ratio and time–work questions throughout the paper.

No. Every SSC power-remainder question yields to cycle-hunting — multiply the base a few times until the remainder hits 1 or −1, then reduce the exponent. The cycle is almost always 6 steps or fewer.

Split the divisor into coprime factors and apply both simple rules: 72 = 8 × 9 (last three digits + digit sum), 88 = 8 × 11, 132 = 4 × 3 × 11. The split must be coprime, or the test passes numbers it shouldn't.

Events repeating together (bells, lights, laps) → LCM of intervals. A greatest number dividing several numbers → HCF. Same remainder in each case → HCF of differences. Least number leaving remainder r with several divisors → LCM + r.

Factors pair up as (d, N/d). The pairs are distinct except when d = N/d, i.e., d = √N — which exists only for perfect squares, leaving one unpaired factor and hence an odd count.