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: . The factorisation answers most integer questions directly:
- Divisibility. is divisible by when every prime power in also appears in . So is divisible by , but not by .
- 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 and , 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 and , the GCD is and the LCM is . For any two positive integers,
and here . LCM questions often hide inside schedules: events that repeat every 18 minutes and every 24 minutes coincide every minutes.
Remainders
Dividing by gives with . The remainder is always smaller than the divisor, and it can be 0, which means divides . The smallest non-negative number with a given remainder is the remainder itself: if leaves remainder 5 on division by 7, 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 even, odd odd | even |
| odd 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 consecutive integers is always divisible by : contains a multiple of 2 and a multiple of 3, so it is divisible by 6 whatever 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 for which is a perfect square?
Factorise in small steps: .
A divisor is with from 0 to 2, from 0 to 2 and from 0 to 1, so there are divisors.
Check by pairing divisors whose product is 252: , , , , , , , , . Nine pairs give 18 divisors.
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: . Check: .
Answer
252 has 18 positive divisors, and . Multiplying by 252 itself also gives a square, but 252 is not the smallest such .
Example 2: Remainder of an expression
When the positive integer is divided by 7, the remainder is 5. What is the remainder when is divided by 7?
Write what the condition means: for some whole number .
Substitute: , so the remainder is 2.
Second route: replace by its remainder. , and leaves 2. Testing gives , 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 and have greatest common divisor 6 and least common multiple 90. If , what is ?
Use : , so .
Check with prime powers. , and .
For 2, the GCD and LCM both have exponent 1, so has . For 3, already supplies the larger exponent 2, so must supply the smaller one, . For 5, has none, so must have . Hence .
Answer
. Picking 90 because it is the LCM fails: , 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 . 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 is divided by 8, the remainder is 5. What is the remainder when is divided by 8?
- A.0
- B.1
- C.5
- D.7
- E.8
Show the answer
Answer: A. 0
Working
Replace by its remainder: , so the remainder is 0. Algebra confirms it for every such : with , , a multiple of 8. Test: gives .
Why the other options are wrong
- B. This is the remainder of alone (); the term was dropped.
- C. This repeats the remainder of , or gives the quotient ; neither is the remainder of the expression.
- D. This is the remainder of alone (); the 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 is the smallest positive integer for which is the cube of an integer, what is ?
- A.6
- B.15
- C.30
- D.45
- E.90
Show the answer
Answer: D. 45
Working
Factorise: . 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 , and .
Why the other options are wrong
- A. This makes a perfect square, , not a cube.
- B. Adding one to each short exponent gives , and is still not a cube.
- C. 30 is the cube root of , not itself.
- E. This includes an unneeded 2: is already a cube, and 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 and are integers and is odd, which of the following must be true?
- A. is odd.
- B. is even.
- C. is odd.
- D. is odd.
- E. is even.
Show the answer
Answer: C. is odd.
Working
A product is odd only when every factor is odd. So is odd, and is odd, which makes even and therefore even. An even number plus an odd number is odd, so is odd. Check with , : is odd and is odd.
Why the other options are wrong
- A. If were odd, would be even and the whole product even.
- B. An even would make the product even.
- D. has to be even, so is even.
- E. is even and 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.