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Question Number 119035 by ZiYangLee last updated on 21/Oct/20

$$\mathrm{Among}\mathrm{the}\mathrm{positive}\mathrm{integers}\mathrm{less}\mathrm{than}1200,$$$$\mathrm{how}\mathrm{many}\mathrm{of}\mathrm{them}\mathrm{are}\mathrm{relatively}\mathrm{prime}$$$$\mathrm{to}60?$$

Answered by floor(10²Eta[1]) last updated on 21/Oct/20

$$60=2^2.3.5$$$$\mathrm{so}\mathrm{we}\mathrm{want}\mathrm{to}\mathrm{know}\mathrm{how}\mathrm{many}\mathrm{numbers}$$$$\mathrm{that}{\mathrm{doesn}}^{\prime }\mathrm{t}\mathrm{have}2,3\mathrm{or}5\mathrm{as}\mathrm{factors}$$$$1200-\lfloor \frac{1200}{2}\rfloor =600\mathrm{numbers}\mathrm{that}\mathrm{are}\mathrm{not}$$$$\mathrm{divisible}\mathrm{by}2\mathrm{from}1\mathrm{to}1200$$$$600-\lfloor \frac{600}{3}\rfloor =400\mathrm{numbers}\mathrm{that}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{have}$$$$\mathrm{factor}3\mathrm{or}2\mathrm{from}1\mathrm{to}1200$$$$400-\lfloor \frac{400}{5}\rfloor =320\mathrm{numbers}\mathrm{that}{\mathrm{doesn}}^{\prime }\mathrm{t}$$$$\mathrm{have}\mathrm{factors}2,3\mathrm{or}5\mathrm{from}1\mathrm{to}1200$$$$\mathrm{Answer}:320$$

Commented by ZiYangLee last updated on 22/Oct/20

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