1.

P Is An Integer. P>883. If P-7 Is A Multiple Of 11, Then The Largest Number That Will Always Divide (p+4) (p+15) Is?

Answer»

<P>GIVEN P is an INTEGER>883.

P-7 is a multiple of 11=>there exist a positive integer a such that

P-7=11 a=>P=11 a+7

(P+4)(P+15)=(11 a+7+4)(11 a+7+15)

=(11 a+11)(11 a+22)

=121(a+1)(a+2)

As a is a positive integer therefore (a+1)(a+2) is divisible by 2.Hence (P+4)(P+15) is divisible by 121*2=242.

Given P is an integer>883.

P-7 is a multiple of 11=>there exist a positive integer a such that

P-7=11 a=>P=11 a+7

(P+4)(P+15)=(11 a+7+4)(11 a+7+15)

=(11 a+11)(11 a+22)

=121(a+1)(a+2)

As a is a positive integer therefore (a+1)(a+2) is divisible by 2.Hence (P+4)(P+15) is divisible by 121*2=242.


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