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Tricky Customer

Age 11 to 14 Short
Challenge Level

Answer: 27


Listing the mutliples
Multiples of 3:            (99),      102,        105,        ... ,        150
Position relative to 99:(99), 99 + 1$\times$3, 99 + 2$\times$3, ..., 99 + 51 = 99 + 17$\times$3
Count:                                       1,              2,      ... ,                        17



Multiples of 5:               100,       105,           110,       ... ,       150
Position relative to 100: 100, 100 + 1$\times$5, 100 + 2$\times$5, ... , 100 + 10$\times$5
Count:                                           1,               2,      ... ,           10
                                    11


17 + 11 = 28 numbers not allowed
But some numbers (like 105 and 150) are in both lists! They are multiples of 3 and 5 ie multiples of 15

105, 120, 135, 150 are counted twice

28$-$4 = 24 numbers not allowed
Out of 51 houses altogether
51$-$24 = 27 houses allowed.


Counting the multiples
There are $51$ houses numbered from $100$ to $150$ inclusive. Of these, $17$ are multiples of $3$, eleven are multiples of $5$ and four are multiples of both $3$ and $5$. So the number of houses Charlie can choose from is $$51-(17+11-4) = 27.$$


This problem is taken from the UKMT Mathematical Challenges.
You can find more short problems, arranged by curriculum topic, in our short problems collection.