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$$A$$ and $$B$$ are two multiples of 36, and $$Q$$ is the set of consecutive integers between $$A$$ and $$B$$, inclusive. If $$Q$$ contains 9 multiples of 9, how many multiples of 4 are there in $$Q$$?
__Alternative Method__: __Plug In__ a good number for $$A$$ such as 36. Count 9 multiples of 9 to find $$B$$. Remember that $$A$$ and $$B$$ are also multiples of 9: >$$\color{red}{A=36}$$, $$45$$, $$54$$, $$63$$, $$72$$, $$81$$, $$90$$, $$99$$, $$\color{red}{B=108}$$ Then use the same method to find the number of multiples of 4 between 36 and 108 inclusive: 1. Subtract the relevant extremes: >>$$B-A=108-36=72$$ 2. Divide by 4: >>$$\displaystyle \frac{72}{4}=18$$ 3. Add 1: >>$$18+1=19$$
Correct. [[snippet]] To find the multiples of 4 between $$A$$ and $$B$$, you must subtract the relevant extremes, divide by 4, and add 1. Note that since $$A$$ and $$B$$ are multiples of 36, they are also multiples of 9 and 4, so $$A$$ and $$B$$ serve as the relevant extremes for all cases. The question states that the number of multiples of 9 between $$A$$ and $$B$$ is 9. Use this information and the method of finding the number of multiples of 9 to find $$B-A$$. Subtracting the extremes in the range ($$B-A$$), dividing by 9, and adding 1 should yield a result of 9. >$$\displaystyle \frac{B-A}{9} + 1 = 9$$ >$$\displaystyle \frac{B-A}{9} = 8$$ Therefore, $$B-A = 8\times 9 = 72$$. Now, how do you find the multiples of 4 that are in that range? Do the same, but now divide by 4: >$$\displaystyle \frac{B-A}{4} + 1 = \frac{72}{4} + 1 = 18 + 1 = 19$$