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 α and β , such that p α and p + 2 β divide Δ , is _______.

Question

Let p and p + 2 be prime numbers and let &#916; = 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 &#945; and &#946; , such that p &#945; and p + 2 &#946; divide &#916; , is _______.

Quantrex Academy · Accepted Answer

Correct answer: 0

Given, &#916; = P ! P + 1 ! P + 2 ! P + 1 ! P + 2 ! P + 3 ! P + 2 ! P + 3 ! P + 4 ! Now using the factorial concept and taking common terms we get, &#8658; &#916; = P ! P + 1 ! P + 2 ! 1 1 1 P + 1 P + 2 P + 3 P + 2 P + 1 P + 3 P + 2 P + 4 P + 3 Now on solving determinant we get, 1 1 1 P + 1 P + 2 P + 3 P + 2 P + 1 P + 3 P + 2 P + 4 P + 3 Using operation C 1 &#8594; C 1 - C 2 &#160; &#38; &#160; C 2 &#8594; C 2 - C 3 0 0 1 - 1 - 1 P + 3 - 2 P - 4 - 2 P - 6 P + 4 P + 3 = 2 P + 6 - 2 P + 4 = 2 Now putting the value of

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 α and β , such that p α and p + 2 β divide Δ , is _______.

Correct answer

Step-by-step solution

More from Determinants

Let p and p + 2 be prime numbers and let &#916; = 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 &#945; and &#946; , such that p &#945; and p + 2 &#946; divide &#916; , is _______.

Correct answer

Step-by-step solution

More from Determinants

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 α and β , such that p α and p + 2 β divide Δ , is _______.