Remainder Theorem Tests — Pick a Level and Start
Tap a level to see every test in it. Start at Level 1 the day you finish the rules, then move to Level 2 with the timer running to push accuracy under pressure.
Level 1 Basic Checking… View tests
One rule applied cleanly — the remainder of a product by replacing each factor, a power whose base is one less than the divisor, a simple cycle, a successive-division chain rebuilt backwards, and f(a) for a linear divisor. Attempt these without the clock while the rules are still fresh.
Level 2 Tier I Standard Checking… View tests
Real SSC CGL Tier I difficulty — two ideas chained in one question: reduce to a factor divisor and then apply a polynomial rule, two bases with two different cycles added together, an unknown divisor recovered before a second remainder is asked for, and coefficients fixed by exact divisibility before a fresh remainder is computed. This is the score that tells you the chapter is exam-ready.
Remainder questions are among the most reliable marks in the SSC CGL Quantitative Aptitude paper, because almost none of them ask you to actually divide. A shift usually carries one that belongs to the chapter outright — a large power by a small divisor, a product of three numbers, or a polynomial divided by a linear factor — and the same reasoning shows up again inside divisibility, unit-digit and LCM–HCF questions.
These practice tests put twenty-five remainder questions in front of you in one sitting, which is what turns the rules from something you can recite into something you spot at a glance. Every question comes with a full solution that names the rule it used, plus the shortcut, in English or Hindi.
How Many Remainder Questions Are Asked in SSC CGL?
The chapter usually gets one to two direct questions, and its methods are reused inside divisibility, unit-digit and number-system questions across the arithmetic block.
| Chapter | Tier I (out of 25) | Tier II Paper I (out of 30) |
|---|---|---|
| Remainder Theorem | 1–2 | 1–3 |
| Number System | 2–3 | 3–4 |
| Rule of Divisibility | 1–2 | 1–2 |
| LCM & HCF | 1–2 | 1–2 |
| Decimal & Fraction | 1–2 | 1–2 |
Typical spread across recent SSC CGL shifts. SSC does not publish a chapter-wise breakup, so treat these as planning ranges. Tier I maths is always 25 questions in a locked 15-minute section.
What You Will Practice in the Remainder Theorem Chapter
Core topics
- The dividend formula, and rebuilding a number from divisor, quotient and remainder
- Remainders of sums, differences and products by replacing each term — including negative remainders
- Powers whose base is one more or one less than the divisor, decided by the parity of the exponent
- Cyclicity — finding the smallest power that leaves 1 or −1 and reducing the exponent by it
- Sums of consecutive powers, where terms repeat or cancel in pairs
- Expressions in n when only the remainder of n is known, and passing to a factor of the divisor
- Recovering an unknown divisor from two remainder facts
- Successive division — rebuilding the number, reversing the order, and least such numbers
- Numbers satisfying two or three remainder conditions at once
- Divisibility of an ± bn and “always divisible by” expressions
- The polynomial remainder theorem — f(a), exact divisibility, and what to add or subtract
- Long repeated-digit numbers, factorial sums and power towers
Power rules you must know cold
| Form | Remainder by d | Example |
|---|---|---|
| (d + 1)n | 1, for every n | 9100 by 8 → 1 |
| (d − 1)n, n even | 1 | 3564 by 36 → 1 |
| (d − 1)n, n odd | d − 1 | 2345 by 24 → 23 |
| ak leaves 1 | reduce n by k | 33 = 27 → 1 (mod 13), so 3100 → 3 |
| ak leaves −1 | a2k leaves 1 | 34 = 81 → −1 (mod 41), so 350 → 9 |
| Base bigger than d | reduce the base first | 10055 by 7 = 255 → 2 |
The exponent is always reduced by the cycle length, never by the divisor. Regrouping often creates a base of d ± 1: 233 + 1 = 811 + 1, and 8 leaves −1 by 9, so the whole thing is divisible by 9.
Divisibility of an ± bn
| Expression | Always divisible by | Example |
|---|---|---|
| an − bn | a − b, for every n | 15n − 4n by 11 |
| an + bn, n odd | a + b | 1125 + 525 by 16 |
| an − bn, n even | a + b as well | 132n − 52n by 144 |
| ak − 1 with k dividing n | divides an − 1 | 25 − 1 = 31 divides 230 − 1 |
“Largest number that always divides” means test the smallest case: 152n − 92n at n = 1 is exactly 144, so 288 is ruled out even though 144 divides every later value.
Also worth memorising
- Dividend = Divisor × Quotient + Remainder, and the remainder is always smaller than the divisor
- If f is a factor of d, the remainder by f is just the old remainder reduced by f
- Unknown divisor: d divides kr − r′, and d is larger than every remainder shown
- Successive remainders r1, r2 by d1, d2 give remainder r1 + d1r2 by d1d2
- Two co-prime conditions repeat every product of the divisors; when each remainder is one less than its divisor, N + 1 is a common multiple
- k consecutive numbers are divisible by k!; for odd k, k2 − 1 is a multiple of 8 and k3 − k of 24; n5 − n is always a multiple of 30
- By 9 use the digit sum, by 11 the alternating sum, by 8 the last three digits, and 1001 = 7 × 11 × 13 kills any block of six equal digits
- n! is a multiple of d once it contains all the prime factors of d — of 24 from 4!, of 5 from 5!, but of 25 only from 10!
Solved Examples — SSC CGL Remainder Theorem
How the Remainder Theorem Chapter Tests Work
- 25 questions per test, with a full question palette — answered, skipped, marked for review, not visited.
- Optional 30-minute timer — with the clock to build exam speed, without it while you are still learning the rules.
- Marking: +2 correct, −0.5 wrong, 0 unattempted — the same scheme as the real paper.
- Solutions, shortcut tricks and concept notes open after you submit, question by question; every solution names the rule it used.
- English and हिंदी, chosen at the start of every test.
- Free with a TrickySSC account — sign in once with Google or your mobile number, and every test is open to you.
How to Prepare Remainder Theorem for SSC CGL
- Start with the base ± 1 rules. Most exam powers are engineered so the base sits next to the divisor; once you check that first, half the chapter answers itself in one line.
- Learn to use negative remainders. Writing 49 as −1 by 50 turns a four-factor product into one multiplication. Just count the minus signs at the end.
- Find the cycle by hand, not by a named theorem. Compute a, a2, a3 … until you hit 1 or −1. SSC never needs anything beyond that.
- Reduce after every step. A remainder bigger than the divisor means the question is not finished — this single check removes most wrong ticks in the chapter.
- For unknown divisors, list the divisors and then apply the bound. "The first number above the remainder" is not the answer; only a divisor of kr − r′ can be.
- Do Level 1 untimed, the same day you finish the rules. The aim is correct recall, not speed.
- Move to Level 2 with the 30-minute clock on. If a power takes more than thirty seconds, you are still deriving the rule instead of applying it.
- Study it with the neighbouring chapters. Do number system and divisibility alongside — the three share their tricks.
- Verify against a real paper. Once Level 2 is consistent, attempt a previous year paper and then a full mock test under locked sectional timing.
SSC CGL शेषफल प्रमेय (Remainder Theorem) चैप्टर टेस्ट — हिंदी में
शेषफल SSC CGL गणित का सबसे भरोसेमंद अंक देने वाला अध्याय है, क्योंकि इसके लगभग किसी प्रश्न में वास्तव में भाग नहीं देना पड़ता। लगभग हर शिफ्ट में इसका 1–2 प्रश्न सीधे आता है — किसी बड़ी घात का छोटे भाजक से शेषफल, तीन-चार संख्याओं के गुणनफल का शेषफल, या किसी बहुपद को रैखिक गुणनखंड से भाग देना — और यही तर्क विभाज्यता, इकाई अंक और ल.स./म.स. के प्रश्नों में भी काम आता है।
TrickySSC पर इस अध्याय के सभी टेस्ट हिंदी में उपलब्ध हैं — प्रश्न, विकल्प, विस्तृत हल और शॉर्टकट ट्रिक सब हिंदी में। दो स्तर हैं: लेवल 1 (आधारभूत) और लेवल 2 (टियर I स्तर)। प्रत्येक टेस्ट में 25 प्रश्न, वैकल्पिक 30 मिनट का टाइमर और +2 / −0.5 की वही मार्किंग जो असली परीक्षा में होती है।
याद रखने योग्य नियम: यदि आधार भाजक से एक अधिक हो तो हर घात का शेषफल 1 होता है; एक कम हो तो सम घात पर 1 और विषम घात पर (भाजक − 1) — जैसे 2345 का 24 से शेषफल 23। किसी और आधार के लिए वह सबसे छोटी घात ढूँढें जो 1 (या −1) छोड़ती है और घातांक को उसी चक्र-लंबाई से कम करें। बहुपद के लिए भाजक का मूल रखकर f(a) निकालें।
पूरे नियम, हल किए उदाहरण और ट्रैप हिंदी नोट्स में देखें: शेषफल प्रमेय स्टडी नोट्स (हिंदी)।