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The-number-of-possible-pairs-of-successive-prime-numbers-such-that-each-of-them-is-greater-than-40-and-their-sum-is-atmost-100-is-




Question Number 14572 by chux last updated on 02/Jun/17
The number of possible pairs of   successive prime numbers such that  each of them is greater than 40 and   their sum is atmost 100 is
$$\mathrm{The}\:\mathrm{number}\:\mathrm{of}\:\mathrm{possible}\:\mathrm{pairs}\:\mathrm{of}\: \\ $$$$\mathrm{successive}\:\mathrm{prime}\:\mathrm{numbers}\:\mathrm{such}\:\mathrm{that} \\ $$$$\mathrm{each}\:\mathrm{of}\:\mathrm{them}\:\mathrm{is}\:\mathrm{greater}\:\mathrm{than}\:\mathrm{40}\:\mathrm{and}\: \\ $$$$\mathrm{their}\:\mathrm{sum}\:\mathrm{is}\:\mathrm{atmost}\:\mathrm{100}\:\mathrm{is} \\ $$
Commented by mrW1 last updated on 02/Jun/17
53+59=112  47+53=100  43+47=90
$$\mathrm{53}+\mathrm{59}=\mathrm{112} \\ $$$$\mathrm{47}+\mathrm{53}=\mathrm{100} \\ $$$$\mathrm{43}+\mathrm{47}=\mathrm{90} \\ $$

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