# How many multiples of 3 are between 10 and 100? (SAT math question)

In the figure above, circular region A represents all integers from 10 to 100, inclusive; circular region B represents all integers that are multiples of 3; and circular region C represents all squares of integers. How many numbers are represented by the shaded region?

• a) 24
• b) 25
• c) 26
• d) 27
• e) 28

Here's my train of thought: All we need to know is this set of numbers is from 10-100 (which I'm assuming does include 10 and 100?) and is multiples of 3. Instead of listing out every multiple of 3, I can have the largest multiple of 3 less than or equal to 100, which is 99, subtract the smallest multiple of 3 more than or equal to 10, which is 12. 99 - 12 = 87 Then to find the number of times 3 fits into 87, divide 87 by 3. 87 / 3 = 29 Agh! 29 is not one of the answer options provided!

Assuming what I've done so far is correct, now I'm thinking that for some reason 1 has to be subtracted from 29 29 - 1 = 28 to get 28, which is one of the options provided and also is the correct answer.

But why would I need to subtract 1 from 29? (Or is that not the right way to find the answer?)

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Note that you are trying to find all multiples of three which are between 10 and 100 WHICH ARE NOT SQUARES OF ANY INTEGER. You must therefore subtract out the number of square between 10 and 100 which are multiples of three... how many are there? Also, you have slightly miscalculated to obtain 29... see lab bhattacharjee's answer. –  AWertheim Oct 15 '13 at 3:22
Oooh, I had ignored the information about the squares. So this is how to rework the problem: 99 - 12 = 87. 87 / 3 = 29. Add extra 1 to count the "other end." 29 + 1 = 30. 36 and 81 are multiples of 3 but are also squares, so we don't include them. 30 - 2 = 28. Great, thanks! –  anita Oct 15 '13 at 3:33

There are $30$ multiples of three between $10$ and $100$ (and it doesn't matter whether we include $10$ and $100$). You are correct that they range from $12$ through $99$, but you made a fencepost error by not adding one. So there are $30$ numbers in the shaded part plus the center of the diagram. The center has numbers that are between $10$ and $100$, squares, and multiples of $3$ (and hence $9$ as they are squares). These are $36$ and $81$. Subtracting those two gets you to $28$.

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$$\left\lfloor\frac{100}3\right\rfloor-\left\lfloor\frac{10}3\right\rfloor$$

Alternatively, the first term is $12,$ the last being $99$ forming an Arithmetic Series with Common Difference $=3$

So, the $n$th term $=99=12+(n-1)3$

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There are $33$ (positive) multiples of $3$ less than $100$, but we need to throw out $3$, $6$, and $9$, so that takes us down to $30$. However, the problem also wants us to throw out any squares that are multiples of $3$, which means square multiples of $9$, which means $9$, $36$, and $81$. We've already thrown out $9$, so we need only throw out the other two, leaving the count at $28$.

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