)135 2that anyaldosititeod wyger is ofthe fprm6q +1, or 6g+ 3, or60+5,

1.	)135 2that anyaldosititeod wyger is ofthe fprm6q +1, or 6g+ 3, or60+5,whereqis some integer.
Answer» Let a be any positive integer and b = 6. Then, by Euclids algorithm, a = 6q + rfor some integer q ≥ 0, and r = 0, 1, 2, 3, 4, 5 because 0 ≤ r Therefore, a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5 Also, 6q + 1 = 2 × 3q + 1 = 2k1+ 1, where k1is a positive integer 6q + 3 = (6q + 2) + 1 = 2 (3q + 1) + 1 = 2k2+ 1, where k2is an integer 6q + 5 = (6q + 4) + 1 = 2 (3q + 2) + 1 = 2k3+ 1, where k3is an integer Clearly, 6q + 1, 6q + 3, 6q + 5 are of the form 2k + 1, where k is an integer. Therefore, 6q + 1, 6q + 3, 6q + 5 are not exactly divisible by 2. Hence, these expressions of numbers are odd numbers. And therefore, any odd integer can be expressed in the form 6q + 1, or 6q + 3, or 6q + 5.

)135 2that anyaldosititeod wyger is ofthe fprm6q +1, or 6g+ 3, or60+5,whereqis some integer.

Answer»

Let a be any positive integer and b = 6.

Then, by Euclids algorithm,

a = 6q + rfor some integer q ≥ 0, and r = 0, 1, 2, 3, 4, 5 because 0 ≤ r

Therefore, a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5

Also, 6q + 1 = 2 × 3q + 1 = 2k1+ 1, where k1is a positive integer

6q + 3 = (6q + 2) + 1 = 2 (3q + 1) + 1 = 2k2+ 1, where k2is an integer

6q + 5 = (6q + 4) + 1 = 2 (3q + 2) + 1 = 2k3+ 1, where k3is an integer

Clearly, 6q + 1, 6q + 3, 6q + 5 are of the form 2k + 1, where k is an integer.

Therefore, 6q + 1, 6q + 3, 6q + 5 are not exactly divisible by 2. Hence, these expressions of numbers are odd numbers.

And therefore, any odd integer can be expressed in the form 6q + 1, or 6q + 3, or 6q + 5.