GRE Quant · Study & Practice

Remainders & Modular Reasoning

AreaArithmetic DifficultyCore GRE weightageMedium–High — remainder and units-digit patterns appear in number-property and Quantitative Comparison sets

Remainders look intimidating because the basic on-screen calculator will not hand you one directly — it only gives a decimal. That gap is exactly why the GRE likes them: the questions reward reasoning over button-pressing. The core identity is simple, dividend = divisor × quotient + remainder, with the remainder always between 0 and one less than the divisor. From there, a few pattern tools — the cyclic behaviour of units digits, and how remainders combine under addition and multiplication — let you answer "what is the units digit of 7⁴⁵?" or "what is the remainder when a huge product is divided by 5?" in seconds. This chapter turns those patterns into reflexes.

Topics

⚡ GRE shortcuts & speed methods

The fastest ways to crack this chapter under time pressure — the techniques that separate a 95+ percentiler from the rest.

  • The calculator gives a decimal, not a remainder. Convert: remainder = N − divisor × (whole-number quotient).
  • Reduce before you compute. Replace each number by its remainder first, then add or multiply — it keeps everything mentally small.
  • Units-digit cycles are all length 1, 2 or 4. Take the exponent mod the cycle length and read the position.
  • Use negative remainders to simplify: 2 ≡ −1 (mod 3) turns 2^n into (−1)^n instantly.
  • Same remainder for several divisors ⇒ answer is LCM(divisors)·k + r.
  • Remainders that are all one short of their divisor ⇒ answer is LCM(divisors)·k − 1.

⚠️ Common mistakes & traps

GRE is designed so that careless errors here cost you marks. Internalise each trap before the exam.

  • Reporting a remainder equal to or larger than the divisor — it must be strictly less.
  • Reading the calculator's decimal part as the remainder (0.4 is not remainder 4).
  • Multiplying big numbers before reducing, instead of reducing to remainders first.
  • Using the wrong cycle length for units digits (assuming everything is a 4-cycle; 4 and 9 are 2-cycles).
  • Treating "remainder 0" as "no remainder / undefined" — remainder 0 just means it divides evenly.
  • Forgetting the shared-remainder shortcut and grinding through lists when LCM·k + r would answer instantly.

📈 GRE exam insight & question patterns

Quantitative Comparison — Quantity A: the units digit of 3⁴⁴. Quantity B: the units digit of 2⁴⁴. Which is greater?

3 cycles 3,9,7,1: 44 ≡ 0 mod 4 → 1. 2 cycles 2,4,8,6: 44 ≡ 0 → 6. Quantity B is greater.

Numeric Entry — enter the remainder when 25 × 33 × 41 is divided by 4.

Reduce mod 4: 25 ≡ 1, 33 ≡ 1, 41 ≡ 1, so the product ≡ 1. Remainder is 1.

A number leaves the same remainder 1 when divided by 2, 3, 4 and 6. What is the smallest such number above 1?

LCM(2,3,4,6) = 12, so it is 12k + 1; the smallest above 1 is 13.

Multiple-answer — which of 6, 10, 16, 22, 28 leave remainder 4 on division by 6? Select all.

16 = 6·2+4 and 22 = 6·3+4 and 28 = 6·4+4 all leave 4; 6 leaves 0 and 10 leaves 4 as well (10 = 6·1+4). So 10, 16, 22, 28.

🎴 Flashcards — instant recall

Tap a card to reveal the answer. Drill these until they are automatic.

Division identityTap to reveal
N = d·q + r, with 0 ≤ r < d
Range of a remainderTap to reveal
0 to (divisor − 1)
Units-digit cycle of 2Tap to reveal
2, 4, 8, 6 (length 4)
Units-digit cycle of 3Tap to reveal
3, 9, 7, 1 (length 4)
Digits with no cycleTap to reveal
0, 1, 5, 6 (stay the same)
Remainder of a productTap to reveal
multiply the remainders, then reduce
Same remainder r for several divisorsTap to reveal
LCM(divisors)·k + r
2 mod 3 as a negativeTap to reveal
−1, so 2^n ≡ (−1)^n

📌 Quick revision

  • Every division obeys N = d·q + r with the remainder strictly less than the divisor.
  • The basic calculator gives decimals, not remainders — convert deliberately.
  • Units digits of powers cycle in lengths 1, 2 or 4; reduce the exponent mod the cycle length.
  • Remainders add and multiply, so reduce numbers to their remainders before computing.
  • Negative remainders (like 2 ≡ −1 mod 3) collapse large powers instantly.
  • A shared remainder across divisors gives LCM·k + r; all-one-short gives LCM·k − 1.
  • Divisor–remainder word problems reduce to the division identity plus an LCM.
  • Remainder 0 is legitimate — it simply means the divisor goes in evenly.

Chapter test

🏆 Vidaara GRE success checklist

You have truly mastered Remainders & Modular Reasoning when you can tick every box below.

  • Recall every formula in this chapter without looking them up
  • Solve each topic’s practice set with at least 80% accuracy
  • Use the chapter shortcuts to cut your solving time in half
  • Spot and avoid every common trap listed above
  • Score 80%+ on the timed chapter test

📋 Chapter mastery scorecard

Track where you stand. Aim for the target before moving to the next chapter.

Skill checkpointTarget
Concept theory & formulas understood100%
Topic practice sets attempted (4 topics)4/4
Best topic-test score— → 80%+
Chapter test score— → 80%+
Flashcards drilled to instant recall8 cards