Consecutive numbers are a set of numbers in which each member of the set is the successor of its predecessor.
4, 6, 8, 10
even consecutive numbers
3, 5, 7, 9
odd consecutive numbers
3, 5, 7, 11, 13
prime consecutive numbers:
Beware of "inclusive questions"
When the GRE says "inclusive" it means that they are counting the first and the last of a given set. Sometimes the GRE doesn't even say "inclusive" and it is implied in the question.
How many numbers are there from 2 to 6?
Normally you think to subtract. 6 - 2 = 4. But, try counting the numbers from 2 to 6. 2, 3, 4, 5, 6. The answer is 5.
Example
A fence consists of a fence post (1 foot wide) and a bridge connecting each post that is 9 feet long. How many fence posts would be required for a 100 foot fence?
Solution
The trick here is to realize that at the end you have an extra post. This means that there would be a total of 10 posts, plus the final one (11 posts).
Sum of Consecutive Integers
What is the sum of the numbers from 1 to 100?
Now, if you are taking the GRE and scribbling on your dry erase board numbers from 1 to 100, it means that you are wasting your time! Surely the GRE would never have you do such labor intensive mathematics. So how could you solve such an onerous problem?
How to solve these questions:
In general to find the sum of all the numbers from F to L (where F is the first number and L is the last number), use the formula:
Consecutive Number Magic Tricks (NOTE: ADVANCED CONTENT - lower scorers consider skipping this section and coming back to it later).
Consecutive Number Magic Trick #1 If you have an ODD (not even) number of numbers in your set, the sum will always be a multiple of the number of numbers.
5 numbers in this set of consecutive numbers:
1 + 2 + 3 + 4 + 5 = 15.
YES, 15 is a multiple of 5.
6 numbers in a set of consecutive numbers:
11 + 12 + 13 + 14 + 15 + 16 = 81.
NO, and there are 6 numbers (an EVEN number of numbers). 81 isn't a multiple of 6.
What is the secret to this trick?
Let's try out the algebra with a set of 5 consecutive integers (an ODD number of numbers).
(x + 1) + (x + 2) + (x + 3) + (x + 4) + (x + 5) = 5x + 15
5x + 15 is a multiple of 5
Consecutive Number Magic Trick #2 If you multiply an ODD number of numbers in a consecutive number set, it WILL ALWAYS be divisible by the number of numbers.
For example, a series of 5 consecutive integers, it WILL ALWAYS be divisible by 5.
2 × 3 × 4 × 5 × 6 = 720
Here there are 5 numbers. Why does this rule work? Because a group of 5 consecutive numbers MUST include a multiple of 5 in it. (2, 3, 4, 5, 6) or (18, 19, 20, 21, 22).
Try this example:
6 numbers in a row. Here we added a 1 in the front.
1 × 2 × 3 × 4 × 5 × 6 = 720
Here there are 6 numbers, and yes, 720 is divisible by 6.
Number Series Number Magic Trick #3
In the PRODUCT of a number series:
if it has one even number (multiple of 2), then the product is a multiple of 2. 6 × 7 = 42
If it has TWO even numbers, then it is a multiple of 4, ALWAYS: 6 × 7 × 8 = 336. Abracadabra! 336 is a multiple of 4 (4 × 84)
If it has THREE even numbers, then it is a multiple of 8, ALWAYS: 6 × 7 × 8 × 9 × 10 = 30240 Abracadabra! 30240 is a multiple of 8 (3780 × 8 = 30240)
Ok, 800score, explain this one! If the product of several numbers contains only ONE even number, the result will be even because anything multiplied by an even number is ALWAYS even. So that one is easy.
The second one: a product of a consecutive number series that contains TWO even numbers must be a multiple of 4. You can figure this one out by doing algebra.
Here is our consecutive numbers series: x × (x + 1) × (x + 2)
If x is even, then (x + 2) must be even as well. Therefore, you have two even numbers as factors. If you prime factorize the number you must have two 2's in it. Therefore, the product of 2 × 2 = 4 MUST also be a divisor (factor) of the number series.
Head spinning? Plug-In numbers:
4 × 5 × 6
6 has 2 as a factor and 4 has 2 as a factor, therefore your end result product must be a multiple of 2 × 2 = 4.
Go one step further: if you have 3 even numbers in a consecutive number series, the product of that series must have three 2's as prime factors. Therefore, the result must have 8 as a factor.
