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Question Number 27057 by Rasheed.Sindhi last updated on 01/Jan/18

$$\mathrm{Try}\mathrm{to}\mathrm{write}\mathrm{new}\mathrm{year}\mathrm{number}$$$$\left(2018\right)\mathrm{as}:$$$$\left(\mathrm{i}\right)\mathrm{Sum}\mathrm{of}\mathrm{two}\mathrm{primes}$$$$\left(\mathrm{ii}\right)\mathrm{Sum}\mathrm{of}\mathrm{three}\mathrm{primes}$$$$\left(\mathrm{iii}\right)\mathrm{Sum}\mathrm{of}\mathrm{primes}$$$$\left(\mathrm{iv}\right)\mathrm{Sum}\mathrm{of}\mathrm{as}\mathrm{many}\mathrm{distinct}\mathrm{primes}\mathrm{as}$$$$\mathrm{possible}.$$

Commented by prakash jain last updated on 01/Jan/18

$$\left(\mathrm{iv}\right)\mathrm{is}\mathrm{easy}\mathrm{for}\mathrm{every}\mathrm{even}\mathrm{number}$$$$2018=2\times 1009.$$$$\mathrm{Amore}\mathrm{interesting}\mathrm{question}\mathrm{can}\mathrm{be}$$$$\max \mathrm{number}\mathrm{of}\mathrm{prime}\mathrm{parts}\mathrm{with}$$$$\mathrm{distinct}\mathrm{primes}.$$

Commented by math solver last updated on 01/Jan/18

$$usesieveofsundaram$$$$and{goldbach}^{\prime }sconjecture!:)$$$$enjoy!$$

Commented by Rasheed.Sindhi last updated on 01/Jan/18

$$\mathrm{Yes}\mathrm{sir}\mathrm{I}\mathrm{actually}\mathrm{mean}\mathrm{distinct}$$$$\mathrm{primes}\mathrm{but}\mathrm{I}{\mathrm{didn}}^{\prime }\mathrm{t}\mathrm{write}\mathrm{the}$$$$\mathrm{condition}\mathrm{mistakenly}.$$

Answered by Rasheed.Sindhi last updated on 02/Jan/18

$$\left(\mathrm{i}\right)7+2011=2018$$$$19+1999=2018$$$$31+1987=2018$$$$2+5+2011=2018$$$$2+17+1999=2018$$

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