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Let p and p + 2 be prime numbers and let

Δ=|p!(p+1)!(p+2)!(p+1)!(p+2)!(p+3)!(p+2)!(p+3)!(p+4)!|

Then the sum, of the maximum values of and , such that p and (p + 2) divide Δ, is…………..

0 2 Views | Posted 2 months ago
Asked by Shiksha User

  • 1 Answer

  • V

    Answered by

    Vishal Baghel | Contributor-Level 10

    2 months ago

    Δ=p!(p+1)!(p+2)!Δ1=2p!(p+1);(p+2)!=2(p!)3.(p+1)2.(p+2)1→α+β=3+1=4

    Δ1=|1p+1(p+2)(p+1)1p+2(p+3)(p+2)1p+3(p+4)(p+3)|=|0−1(p+2)(−2)0−1(p+3)(−2)1p+3(p+4)(p+3)|

    = 2(p + 3) – 2 (p + 2) = 2

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