Consecutive prime numbers

Number Theory Level 2

Suppose that (a,b,p,q) \text{(a,b,p,q)} are consecutive prime numbers such that (a<b<p<q) \text{(a<b<p<q)} .

Is it possible that a+a+b+p+q = k \text{a+a+b+p+q}=k such that k is a prime number \text{k is a prime number} ?

No Yes None Not sufficient information

Matin Naseri by Matin Naseri , 17, Afghanistan

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1 solution

Matin Naseri
Mar 10, 2018

We have (2<3<5<7) \text{(2<3<5<7)} and also all of them are consecutive prime numbers.

2+2+3+5+7=19 \text{2+2+3+5+7=19}

Thus the answer is y e s \boxed{\color{#302B94}{yes}}

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Eg 7, 13, 19, 29 as consecutive prime numbers => 7+7+13+19+29 = 75; 7, 13, 19, 31 as consecutive prime numbers => 7+7+13+19+31 = 77

E Koh - 3 years, 1 month ago

Consider 7 11 13 17 as consecutive prime numbers => 7+7+11+13+17 = 55 Consider 13 17 19 23 as consecutive prime numbers => 13+13+17+19+23 = 85

E Koh - 3 years, 1 month ago

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It´s possible , not mandatory.

Ricardo Moritz Cavalcanti - 10 months, 3 weeks ago

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