Bunuel wrote:
marcusaurelius wrote:
How many multiples of 4 are there between 12 and 96, inclusive?
21
22
23
24
25
My answer was 21 and that's incorrect.
- \(Number \ of \ multiples \ of \ x \ in \ the \ range = \)
\(=\frac{Last \ multiple \ of \ x \ in \ the \ range \ - \ First \ multiple \ of \ x \ in \ the \ range}{x}+1\).
In the original case: \(\frac{96-12}{4}+1=22\).
If the question were: how many multiples of 5 are there between -7 and 35, not inclusive?
Last multiple of 5
IN the range is 30;
First multiple of 5
IN the range is -5;
\(\frac{30-(-5)}{5}+1=8\).
OR:How many multiples of 7 are there between -28 and -1, not inclusive?
Last multiple of 7
IN the range is -7;
First multiple of 7
IN the range is -21;
\(\frac{-7-(-21)}{7}+1=3\).
Hope it helps.
This formula is helpful for many questions regarding sets or sums of sets with consecutively spaced integers. I'm just taking a moment to explain where the formula comes from, because it's something the GMAT tests in a sneaky way.
This value of 'last number' - 'first number' is called the 'range' of a set. That can basically be thought of as the 'distance' from the smallest number to the largest.
Students often have a sense that we should divide that distance by the number we're counting by (in the original question, '4').
The thing that often gets forgotten is the '+1.' Why do we have that plus one?
Well, what do we solve for when we divide 'the range' by 'the number we're counting by?'
For instance, the range from 12 to 96 is 84. If we divide that 84 by the 'four' we're counting by, we get 21. What does that 21 'signify' for this problem???
It tells you how many 'jumps' of four you make to get from 12 to 96.
12 + (4 + 4 + 4 + 4 .... +4) {21 times} = 96
Visualized this might be seen as:
12 _(+4)_ 16 _(+4)_ 20 _(+4)_ 24 .... 92 _(+4)_ 96
So we have 21 of those '+4s'... that is 21 gaps BETWEEN the numbers in the set. In order to have 21 gaps between numbers, you must have 22 numbers total! That's why we must add 1.
POINT IS: Do not mix up the 'range' with the 'number of numbers' (and it's very easy to do when you 'cut' the range into pieces, as we do in this problem).
(Random aside, I often see people try to count up how many multiples of [the number] there are in a range of the first '10' numbers, say, and then multiply that by how many '10s' there are in the whole set. This won't often work---some '10' will have more multiples of that number than other '10s.'
For instance 0-10 has three multiples of 3. 11-20 has three multiples of 3. But 21-30 has *four* multiples of 3. Stuff like that will happen often.