Lesson 2 of 35 · Quantitative Reasoning: arithmetic

Number properties: primes, divisibility, remainders and parity

Use prime factorisation, divisibility rules, remainders and odd-even rules to settle integer questions in a line instead of a page of arithmetic.

About 20 minutes

In this lesson

  • Factorise an integer into primes and read off its divisors, and whether it is a square or a cube.
  • Find a GCD or LCM from prime powers, and combine divisibility conditions correctly.
  • Work out remainders of sums and products without dividing large numbers.
  • Decide what must be true about odd, even and negative integers.

The idea

Why properties beat calculation

GMAT Quantitative Reasoning gives you 45 minutes for 21 problem-solving questions, a little over two minutes each, and you cannot use a calculator. Questions about integers are usually built so that multiplying everything out is slow, while one property of the numbers settles the answer in a line. This lesson collects the properties that do most of that work.

Primes and prime factorisation

A prime has exactly two positive divisors, 1 and itself. So 1 is not prime, and 2 is the only even prime. The primes below 30 are 2, 3, 5, 7, 11, 13, 17, 19, 23 and 29.

Every integer greater than 1 is a product of primes in exactly one way. Break a number down with small factors first: 252=4×63=22×32×7252 = 4 \times 63 = 2^2 \times 3^2 \times 7. The factorisation answers most integer questions directly:

  • Divisibility. nn is divisible by mm when every prime power in mm also appears in nn. So 252252 is divisible by 18=2×3218 = 2 \times 3^2, but not by 8=238 = 2^3.
  • Number of divisors. Add 1 to each exponent and multiply. A divisor chooses, for every prime, an exponent from zero up to the maximum, so the choices multiply.
  • Squares and cubes. A perfect square has only even exponents; a perfect cube has exponents that are all multiples of 3.

Divisibility rules worth knowing

Divisor Test
2, 5, 10 the last digit
4 the last two digits form a multiple of 4
8 the last three digits form a multiple of 8
3, 9 the digit sum is divisible by 3, or by 9
6 divisible by both 2 and 3

To combine conditions, use prime powers, not multiplication. A number divisible by 4 and by 6 must contain 222^2 and 33, so it is divisible by 12; it need not be divisible by 24, and 12 itself shows it.

Greatest common divisor and least common multiple

Write both numbers as prime powers. The GCD takes the smaller exponent of each prime, the LCM the larger. For 72=23×3272 = 2^3 \times 3^2 and 60=22×3×560 = 2^2 \times 3 \times 5, the GCD is 22×3=122^2 \times 3 = 12 and the LCM is 23×32×5=3602^3 \times 3^2 \times 5 = 360. For any two positive integers,

gcd⁡(a,b)×lcm(a,b)=ab\gcd(a, b) \times \text{lcm}(a, b) = ab

and here 12×360=4320=72×6012 \times 360 = 4320 = 72 \times 60. LCM questions often hide inside schedules: events that repeat every 18 minutes and every 24 minutes coincide every lcm(18,24)=72\text{lcm}(18, 24) = 72 minutes.

Remainders

Dividing nn by dd gives n=qd+rn = qd + r with 0≤r<d0 \le r < d. The remainder is always smaller than the divisor, and it can be 0, which means dd divides nn. The smallest non-negative number with a given remainder is the remainder itself: if nn leaves remainder 5 on division by 7, nn could be 5, 12, 19 and so on.

Remainders survive addition and multiplication. To find the remainder of a sum or a product, replace each number by its remainder, compute, and reduce again. That is why you may test a remainder question with the smallest number that fits and trust the result for every other number that fits.

Odd, even and sign

Operation Result
even ±\pm even, odd ±\pm odd even
odd ±\pm even odd
a product with at least one even factor even
a product of odd numbers only odd

Zero is even. A product is negative exactly when it has an odd number of negative factors, and a square is never negative. The product of kk consecutive integers is always divisible by k!k!: n(n+1)(n+2)n(n+1)(n+2) contains a multiple of 2 and a multiple of 3, so it is divisible by 6 whatever nn is.

Testing cases

For "which must be true" questions, try numbers that break careless rules: 0, 1, a negative number, and one even and one odd value. A single counterexample is enough to discard an option; an option survives only if no allowed value breaks it.

Worked examples

Example 1: Divisors and a perfect square from one factorisation

How many positive divisors does 252 have, and what is the smallest positive integer kk for which 252k252k is a perfect square?

  1. Factorise in small steps: 252=4×63=22×32×7252 = 4 \times 63 = 2^2 \times 3^2 \times 7.

  2. A divisor is 2a×3b×7c2^a \times 3^b \times 7^c with aa from 0 to 2, bb from 0 to 2 and cc from 0 to 1, so there are 3×3×2=183 \times 3 \times 2 = 18 divisors.

  3. Check by pairing divisors whose product is 252: 1×2521 \times 252, 2×1262 \times 126, 3×843 \times 84, 4×634 \times 63, 6×426 \times 42, 7×367 \times 36, 9×289 \times 28, 12×2112 \times 21, 14×1814 \times 18. Nine pairs give 18 divisors.

  4. For a square every exponent must be even. The exponents of 2 and 3 already are; 7 has exponent 1, so one more 7 is needed: k=7k = 7. Check: 252×7=1764=422252 \times 7 = 1764 = 42^2.

Answer

252 has 18 positive divisors, and k=7k = 7. Multiplying by 252 itself also gives a square, but 252 is not the smallest such kk.

Example 2: Remainder of an expression

When the positive integer nn is divided by 7, the remainder is 5. What is the remainder when 4n+34n + 3 is divided by 7?

  1. Write what the condition means: n=7q+5n = 7q + 5 for some whole number qq.

  2. Substitute: 4n+3=28q+23=28q+21+2=7(4q+3)+24n + 3 = 28q + 23 = 28q + 21 + 2 = 7(4q + 3) + 2, so the remainder is 2.

  3. Second route: replace nn by its remainder. 4×5+3=234 \times 5 + 3 = 23, and 23=21+223 = 21 + 2 leaves 2. Testing n=12n = 12 gives 51=49+251 = 49 + 2, the same.

Answer

The remainder is 2. Answering 23 forgets to reduce: a remainder on division by 7 must be less than 7.

Example 3: A missing number from its GCD and LCM

Positive integers aa and bb have greatest common divisor 6 and least common multiple 90. If a=18a = 18, what is bb?

  1. Use gcd⁡×lcm=ab\gcd \times \text{lcm} = ab: 6×90=540=18b6 \times 90 = 540 = 18b, so b=30b = 30.

  2. Check with prime powers. 6=2×36 = 2 \times 3, 90=2×32×590 = 2 \times 3^2 \times 5 and a=18=2×32a = 18 = 2 \times 3^2.

  3. For 2, the GCD and LCM both have exponent 1, so bb has 212^1. For 3, aa already supplies the larger exponent 2, so bb must supply the smaller one, 313^1. For 5, aa has none, so bb must have 515^1. Hence b=2×3×5=30b = 2 \times 3 \times 5 = 30.

Answer

b=30b = 30. Picking 90 because it is the LCM fails: gcd⁡(18,90)=18\gcd(18, 90) = 18, not 6.

Common traps

  • Multiplying divisors to combine them

    "Divisible by 4 and by 6" does not mean divisible by 24; it means divisible by lcm(4,6)=12\text{lcm}(4, 6) = 12. The factors share a 2, and multiplying counts it twice. Combine conditions through prime powers, taking the largest exponent of each prime.

  • Testing only comfortable numbers

    "Must be true" options often fail only for 0, 1, a negative number or a fraction the question allows. If the stem says "integer", try a negative and zero; if it says "number", try a fraction too. Check what the stem permits before you test.

  • A remainder that is not reduced

    After replacing numbers by their remainders you may get something larger than the divisor, such as 23 for a division by 7. Reduce again. A remainder is always at least 0 and less than the divisor, so an answer choice equal to or above the divisor can be crossed out at once.

  • Forgetting that 2 is prime and 1 is not

    If two primes add up to an odd number, one of them must be 2, the only even prime. Counting 1 as a prime gives wrong divisor counts and wrong prime lists. Both slips are quick to make under time pressure and quick to avoid if you say the definition to yourself.

Check yourself

Answer each question before you open its answer.

Question 1: Remainder of n² + 3n

Problem Solving

When the positive integer nn is divided by 8, the remainder is 5. What is the remainder when n2+3nn^2 + 3n is divided by 8?

  1. A.0
  2. B.1
  3. C.5
  4. D.7
  5. E.8
Show the answer

Answer: A. 0

Working

Replace nn by its remainder: 52+3×5=40=5×85^2 + 3 \times 5 = 40 = 5 \times 8, so the remainder is 0. Algebra confirms it for every such nn: with n=8q+5n = 8q + 5, n2+3n=n(n+3)=(8q+5)(8q+8)=8(8q+5)(q+1)n^2 + 3n = n(n + 3) = (8q + 5)(8q + 8) = 8(8q + 5)(q + 1), a multiple of 8. Test: n=13n = 13 gives 169+39=208=8×26169 + 39 = 208 = 8 \times 26.

Why the other options are wrong

  • B. This is the remainder of n2n^2 alone (25=24+125 = 24 + 1); the 3n3n term was dropped.
  • C. This repeats the remainder of nn, or gives the quotient 40÷840 \div 8; neither is the remainder of the expression.
  • D. This is the remainder of 3n3n alone (15=8+715 = 8 + 7); the n2n^2 term was dropped.
  • E. Adding the two remainders 1 and 7 gives 8, which must be reduced: a remainder on division by 8 is less than 8.

What carries over

Replace each number by its remainder, compute, then reduce once more.

Question 2: Making a perfect cube

Problem Solving

If kk is the smallest positive integer for which 600k600k is the cube of an integer, what is kk?

  1. A.6
  2. B.15
  3. C.30
  4. D.45
  5. E.90
Show the answer

Answer: D. 45

Working

Factorise: 600=6×100=23×3×52600 = 6 \times 100 = 2^3 \times 3 \times 5^2. Each exponent must reach a multiple of 3: the 2 is fine, the 3 needs two more factors of 3 and the 5 needs one more factor of 5. So k=32×5=45k = 3^2 \times 5 = 45, and 600×45=27 000=303600 \times 45 = 27\,000 = 30^3.

Why the other options are wrong

  • A. This makes 600k600k a perfect square, 3600=6023600 = 60^2, not a cube.
  • B. Adding one to each short exponent gives 600×15=23×32×53600 \times 15 = 2^3 \times 3^2 \times 5^3, and 323^2 is still not a cube.
  • C. 30 is the cube root of 600k600k, not kk itself.
  • E. This includes an unneeded 2: 232^3 is already a cube, and 600×90=24×33×53600 \times 90 = 2^4 \times 3^3 \times 5^3 is not.

What carries over

For a square or a cube, raise every prime exponent to the next multiple of 2 or 3, and add nothing else.

Question 3: Which must be true?

Problem Solving

If xx and yy are integers and y(x2+1)y(x^2 + 1) is odd, which of the following must be true?

  1. A.xx is odd.
  2. B.yy is even.
  3. C.x+yx + y is odd.
  4. D.xyxy is odd.
  5. E.x2+yx^2 + y is even.
Show the answer

Answer: C. x+yx + y is odd.

Working

A product is odd only when every factor is odd. So yy is odd, and x2+1x^2 + 1 is odd, which makes x2x^2 even and therefore xx even. An even number plus an odd number is odd, so x+yx + y is odd. Check with x=2x = 2, y=3y = 3: 3×5=153 \times 5 = 15 is odd and x+y=5x + y = 5 is odd.

Why the other options are wrong

  • A. If xx were odd, x2+1x^2 + 1 would be even and the whole product even.
  • B. An even yy would make the product even.
  • D. xx has to be even, so xyxy is even.
  • E. x2x^2 is even and yy is odd, so the sum is odd.

What carries over

An odd product forces every factor to be odd; start from that and the rest follows.

Key points

  • Prime factorisation answers divisibility, divisor-count, square and cube questions.

  • Combine divisibility conditions with the LCM, never by multiplying the divisors.

  • Replace numbers by their remainders, compute, and reduce below the divisor.

  • An odd product needs every factor odd; zero is even; 2 is the only even prime.

All questions here are original and written for UniExamer; none comes from the test maker.

Written by the UniExamer team with AI assistance. Every worked example and check question was re-solved by an independent checker before publication. No subject expert has reviewed this lesson yet. If you find a mistake, write to support@uniexamer.com.

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