# Show that any positive odd integer is of the form 6q+1, or 6q+3 , or 6q+5, where q is some integer.

Ans.

Let a be any positive integer and b = 6. Then, by Euclid’s algorithm,

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

r < 6.
Therefore, a = 6q or 6q + 1 or 6q +2 or 6q +3 or 6q + 4 or 6q +5
Also, 6q + 1 = 2 x 3q +1 = 2k₁ + 1 , where K₁ is a positive integer
6q+3 = (6q + 2) + 1 = 2 (3q +1 ) +1 + 2k₂ + 1, where k₂ is an integer
6q + 5 = (6q + 4) + 1 = 2 (3q +2) + 1 = 2k₃ + 1 , where k₃ is an integer
Clearly, 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.
Ans.
Let a be any positive integer and b = 6. Then, by Euclid’s algorithm,
a = 6q + r for some integer q ≥ 0, and r = 0, 1, 2, 3, 4, 5 because 0 ≤
r < 6.
Therefore, a = 6q or 6q + 1 or 6q +2 or 6q +3 or 6q + 4 or 6q +5
Also, 6q + 1 = 2 x 3q +1 = 2k₁ + 1 , where K₁ is a positive integer
6q+3 = (6q + 2) + 1 = 2 (3q +1 ) +1 + 2k₂ + 1, where k₂ is an integer
6q + 5 = (6q + 4) + 1 = 2 (3q +2) + 1 = 2k₃ + 1 , where k₃ is an integer
Clearly, 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.
Ans.
Let a be any positive integer and b = 6. Then, by Euclid’s algorithm,
a = 6q + r for some integer q ≥ 0, and r = 0, 1, 2, 3, 4, 5 because 0 ≤
r < 6.
Therefore, a = 6q or 6q + 1 or 6q +2 or 6q +3 or 6q + 4 or 6q +5
Also, 6q + 1 = 2 x 3q +1 = 2k₁ + 1 , where K₁ is a positive integer
6q+3 = (6q + 2) + 1 = 2 (3q +1 ) +1 + 2k₂ + 1, where k₂ is an integer
6q + 5 = (6q + 4) + 1 = 2 (3q +2) + 1 = 2k₃ + 1 , where k₃ is an integer
Clearly, 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.