Show that the cube of a positive integer of the form 6q + r, q is an integer and r = 0, 1, 2, 3, 4, 5 is also of the form 6m+r.

Solution: 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,...

Solution: 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. So, a...

Solution: No, every positive integer cannot be only of the form 4q + 2. Justification: Let a be any positive integer. Then by Euclid’s division lemma, we have a = bq + r, where...

How to assume value of “b”? Here it is assumed as a=4q+0,1,2,3. But in some similar problems value of b is taken as 3. So what is the criteria to assume...

Solution: Let a be the positive integer and b = 5. Then, by Euclid’s algorithm, a = 5m + r for some integer m ≥ 0 and r = 0, 1, 2,...

Solution: Let a be the positive integer and b = 6. Then, by Euclid’s algorithm, a = 6q + r for some integer q ≥ 0 and r = 0, 1, 2,...

By Euclid’s division algorithm a = bq + r, where 0 ≤ r ≤ b Put b = 4 a = 4q + r, where 0 ≤ r ≤ 4 If r = 0,...

Let, n = 6q + 5, when q is a positive integer We know that any positive integer is of the form 3k, or 3k + 1, or 3k + 2 ∴...

Let a be any positive odd integer and b = 6. Then, by Euclid’s algorithm, a = 6q + r, for some integer q ≥ 0 and 0 ≤ r...