Why use a general expression for an integer instead of an example?
GMAC lists number properties, divisibility, even and odd numbers, and remainders among quantitative topics. The examples in this workshop were created to explain these concepts; they are not from an official exam and do not predict how often a topic appears.
Saying n is even means there is an integer k such that n = 2k. An odd integer is written as 2k + 1. These expressions let you reason about every allowed case. Trying n = 4 may help explore an idea, but does not replace this general reasoning.
For division by a positive integer d, write n = dq + r with 0 ≤ r < d. The remainder must satisfy this bound. An expression such as 7q + 9 still needs to be reduced: 7q + 9 = 7(q + 1) + 2, so the remainder when dividing by 7 is 2.
What can you conclude from an odd product?
In a product of integers, one even factor makes the entire product even. Therefore, for the product of two integers to be odd, both factors must be odd.
If a and b are integers and ab is odd, which of the following must be even?
Choose an answer.
The solution, step by step
a + b.
The method in action
- a and b are odd: a = 2u + 1 and b = 2v + 1. Then a + b = 2(u + v + 1), which is even.
- a, b, and ab are odd. Adding 1 to the even sum a + b makes it odd. The third choice is the only one that must be even.
- Trying a = 3 and b = 5 checks the result but does not prove it. The form 2(u + v + 1) establishes the result for all cases.
How can you carry a remainder through an expression?
Replace n with its quotient-remainder form. Multiples of the divisor do not affect the remainder; only a small calculation remains.
When the positive integer n is divided by 7, the remainder is 4. What is the remainder when 3n + 2 is divided by 7?
Choose an answer.
The solution, step by step
0.
The method in action
- n = 7q + 4. So 3n + 2 = 21q + 14 = 7(3q + 2), which is divisible by 7.
- Calculating from the remainder gives 3 × 4 + 2 = 14; 14 cannot be a remainder when dividing by 7. Reducing it gives 0.
- The other choices do not satisfy the general expression. Testing n = 4 and then n = 11 confirms the result but does not replace the factorization.
The pitfall
A conclusion that holds in several trials can still be false
For example, “every odd integer is prime” works for 3, 5, and 7, then fails for 9. Accumulating successful trials proves nothing universal. One allowed counterexample, however, is enough to refute the claim.
For a “must be true” question, look for a reason that covers every case. For “could be true,” one compatible example may be enough. The difference lies in the logical task, not the difficulty of the calculation.
How can you use multiple divisibility conditions?
Being divisible by two numbers does not always mean being divisible by their product. Account for the factors they share.
What is the smallest positive integer divisible by both 12 and 18?
Choose an answer.
The solution, step by step
36.
The method in action
- 12 = 2² × 3 and 18 = 2 × 3². The least common multiple contains 2² and 3²: 4 × 9 = 36.
- 36/12 = 3 and 36/18 = 2. 24 is not divisible by 18; 30 is divisible by neither; 54 is not divisible by 12.
- 216, the product of 12 × 18, is a common multiple but not the least common multiple. Shared factors do not need to be counted twice.
Key takeaways
- Use 2k, 2k + 1, or dq + r, depending on the property.
- Respect the bounds on a remainder.
- A counterexample disproves a rule; successful trials do not prove it.