Question

If m=n2-n, where n is an integer, then m2 -2m is divisible by
A. 20
B. 24
C. 30
D. 16
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Answer 1

Answer:

Any odd positive number is in the form of (4p+1)or(4p+3) for some integer P.

Any odd positive number is in the form of (4p+1)or(4p+3) for some integer P.let

Any odd positive number is in the form of (4p+1)or(4p+3) for some integer P.letn=4p+3

Any odd positive number is in the form of (4p+1)or(4p+3) for some integer P.letn=4p+3n

Any odd positive number is in the form of (4p+1)or(4p+3) for some integer P.letn=4p+3n 2

Any odd positive number is in the form of (4p+1)or(4p+3) for some integer P.letn=4p+3n 2 −1=(4p+1)

Any odd positive number is in the form of (4p+1)or(4p+3) for some integer P.letn=4p+3n 2 −1=(4p+1) 2

Any odd positive number is in the form of (4p+1)or(4p+3) for some integer P.letn=4p+3n 2 −1=(4p+1) 2 −1=16p

Any odd positive number is in the form of (4p+1)or(4p+3) for some integer P.letn=4p+3n 2 −1=(4p+1) 2 −1=16p 2

Any odd positive number is in the form of (4p+1)or(4p+3) for some integer P.letn=4p+3n 2 −1=(4p+1) 2 −1=16p 2 +8p+1−1=8p(2p+1)

Any odd positive number is in the form of (4p+1)or(4p+3) for some integer P.letn=4p+3n 2 −1=(4p+1) 2 −1=16p 2 +8p+1−1=8p(2p+1)⇒n

Any odd positive number is in the form of (4p+1)or(4p+3) for some integer P.letn=4p+3n 2 −1=(4p+1) 2 −1=16p 2 +8p+1−1=8p(2p+1)⇒n 2

Any odd positive number is in the form of (4p+1)or(4p+3) for some integer P.letn=4p+3n 2 −1=(4p+1) 2 −1=16p 2 +8p+1−1=8p(2p+1)⇒n 2 −1isdivisibleby8

Any odd positive number is in the form of (4p+1)or(4p+3) for some integer P.letn=4p+3n 2 −1=(4p+1) 2 −1=16p 2 +8p+1−1=8p(2p+1)⇒n 2 −1isdivisibleby8n

Any odd positive number is in the form of (4p+1)or(4p+3) for some integer P.letn=4p+3n 2 −1=(4p+1) 2 −1=16p 2 +8p+1−1=8p(2p+1)⇒n 2 −1isdivisibleby8n 2

Any odd positive number is in the form of (4p+1)or(4p+3) for some integer P.letn=4p+3n 2 −1=(4p+1) 2 −1=16p 2 +8p+1−1=8p(2p+1)⇒n 2 −1isdivisibleby8n 2 −1=(4p+3)

Any odd positive number is in the form of (4p+1)or(4p+3) for some integer P.letn=4p+3n 2 −1=(4p+1) 2 −1=16p 2 +8p+1−1=8p(2p+1)⇒n 2 −1isdivisibleby8n 2 −1=(4p+3) 2

Any odd positive number is in the form of (4p+1)or(4p+3) for some integer P.letn=4p+3n 2 −1=(4p+1) 2 −1=16p 2 +8p+1−1=8p(2p+1)⇒n 2 −1isdivisibleby8n 2 −1=(4p+3) 2 −1=16p

Any odd positive number is in the form of (4p+1)or(4p+3) for some integer P.letn=4p+3n 2 −1=(4p+1) 2 −1=16p 2 +8p+1−1=8p(2p+1)⇒n 2 −1isdivisibleby8n 2 −1=(4p+3) 2 −1=16p 2

Any odd positive number is in the form of (4p+1)or(4p+3) for some integer P.letn=4p+3n 2 −1=(4p+1) 2 −1=16p 2 +8p+1−1=8p(2p+1)⇒n 2 −1isdivisibleby8n 2 −1=(4p+3) 2 −1=16p 2 +24p+9−1=16p

Any odd positive number is in the form of (4p+1)or(4p+3) for some integer P.letn=4p+3n 2 −1=(4p+1) 2 −1=16p 2 +8p+1−1=8p(2p+1)⇒n 2 −1isdivisibleby8n 2 −1=(4p+3) 2 −1=16p 2 +24p+9−1=16p 2

Any odd positive number is in the form of (4p+1)or(4p+3) for some integer P.letn=4p+3n 2 −1=(4p+1) 2 −1=16p 2 +8p+1−1=8p(2p+1)⇒n 2 −1isdivisibleby8n 2 −1=(4p+3) 2 −1=16p 2 +24p+9−1=16p 2 +24p+8

Any odd positive number is in the form of (4p+1)or(4p+3) for some integer P.letn=4p+3n 2 −1=(4p+1) 2 −1=16p 2 +8p+1−1=8p(2p+1)⇒n 2 −1isdivisibleby8n 2 −1=(4p+3) 2 −1=16p 2 +24p+9−1=16p 2 +24p+8=8(2p

Any odd positive number is in the form of (4p+1)or(4p+3) for some integer P.letn=4p+3n 2 −1=(4p+1) 2 −1=16p 2 +8p+1−1=8p(2p+1)⇒n 2 −1isdivisibleby8n 2 −1=(4p+3) 2 −1=16p 2 +24p+9−1=16p 2 +24p+8=8(2p 2

Any odd positive number is in the form of (4p+1)or(4p+3) for some integer P.letn=4p+3n 2 −1=(4p+1) 2 −1=16p 2 +8p+1−1=8p(2p+1)⇒n 2 −1isdivisibleby8n 2 −1=(4p+3) 2 −1=16p 2 +24p+9−1=16p 2 +24p+8=8(2p 2 +3p+1)

Any odd positive number is in the form of (4p+1)or(4p+3) for some integer P.letn=4p+3n 2 −1=(4p+1) 2 −1=16p 2 +8p+1−1=8p(2p+1)⇒n 2 −1isdivisibleby8n 2 −1=(4p+3) 2 −1=16p 2 +24p+9−1=16p 2 +24p+8=8(2p 2 +3p+1)⇒n

Any odd positive number is in the form of (4p+1)or(4p+3) for some integer P.letn=4p+3n 2 −1=(4p+1) 2 −1=16p 2 +8p+1−1=8p(2p+1)⇒n 2 −1isdivisibleby8n 2 −1=(4p+3) 2 −1=16p 2 +24p+9−1=16p 2 +24p+8=8(2p 2 +3p+1)⇒n 2

• Any odd positive number is in the form of (4p+1)or(4p+3) for some integer P.letn=4p+3n 2 −1=(4p+1) 2 −1=16p 2 +8p+1−1=8p(2p+1)⇒n 2 −1isdivisibleby8n 2 −1=(4p+3) 2 −1=16p 2 +24p+9−1=16p 2 +24p+8=8(2p 2 +3p+1)⇒n 2