State true or false: Any positive odd integer is of the form (6m + 1) or (6m + 3) or (6m + 5), where m is some integer.

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True

False

answer is A.

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Detailed Solution

Given is the statement that any positive odd integer is of the form (6m + 1) or (6m + 3) or (6m + 5), and we have to find if the statement is true or not.
According to the Euclid’s Division Lemma, for any two positive integers say a and b, there exist two unique non-zero integers q and r such that

a = b × q + r

where

0 ≤ r < b

.
And sum of two even numbers is always even but sum of an even and an odd integer is always odd.
Let us assume two integers a and b, where a is some positive integer and b = 6.
So, by Euclid’s division algorithm, we get
a = 6m + r, where r is the remainder and is non zero.
Then we can say that r = 0,1,2,3,4,5 because

0 ≤ r < b

, where r is assumed to be 6.
So, for r = 0, we have, a = 6m, which is an even integer as 6 is even.
Again, for r =1, we have, a = 6m + 1, which is odd as 6 is even but 1 is odd.
Again, for r = 2, we have, a = 6m + 2, which is even as both 6 and 2 are even.
Again, for r = 3, we have, a = 6m + 3, which is odd as 6 is even but 3 is odd.
Again, for r = 4, we have, a = 6m + 4, which is even as 6 and 4 both are even.
Again, for r = 5, we have, a = 6m+5, which is odd as 6 is even but 5 is odd.
So, we observed that the positive odd integers are of the general form (6m + 1), (6m + 3) and (6m + 5).
This affirms the given statement.
Hence, the statement is true.
So, the correct option is 1.

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