Question 309061: prove that the sum of two consecutive multiple of 5 is always an odd number and prove that the product of two consecutive multiple of 5 is always an even number
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! prove that the sum of two consecutive multiple of 5 is always an odd number
1st: 5x where x is an integer
2nd: 5(x+1)
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Sum Form:
5x + 5(x+1)
= 5(x+x+1)
= 5(2x+1)
Since integers are a closed set under multiplcation and addition,
2x+1 is an integer. Therefore the sum is an integer-multiple of 5.
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prove that the product of two consecutive multiple of 5 is always an even number
Product Form:
5x(5(x+1))
Case 1: Assume that x is an odd integer; then x+1 is an even integer.
Let x+1 = 2k.
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= 5x(5(2k)
= 2*(5^2)*kx
Since integers are closed under multiplication, 5^2kx is an integer.
Therefore 5x(5(x+1)) is a multiple of 2 and is therefore an even integer.
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Case 2: Assume x is an even integer.
Then 5x*5(x+1)
= 5(2k)5(x+1)
= 2[5^2*k(x+1)]
By closure of integers and def. of an even integer
the product is an even integer.
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Cheers,
Stan H.
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