Question

Use division algorithm to show that any positive odd integer is of the form of 6q+1 or 6q+3 or 6q+5, where q is some integer.​

Answer 1

Answer:

this is the answer and its

Step-by-step explanation:

According to Euclid’s Division Lemma if we have two positive integers a and b, then there exist unique integers q and r which satisfies the condition a = bq + r where 0 ≤ r < b.

Let a be the positive odd integer which when divided by 6 gives q as quotient and r as remainder.

According to Euclid’s division lemma

a = bq + r

a = 6q + r………………….(1)

where, (0 ≤ r < 6)

So r can be either 0, 1, 2, 3, 4 and 5.

Case 1:

If r = 1, then equation (1) becomes

a = 6q + 1

The Above equation will be always as an odd integer.

Case 2:  

If r = 3, then equation (1) becomes

a = 6q + 3

The Above equation will be always as an odd integer.

Case 3:  

If r = 5, then equation (1) becomes

a = 6q + 5

The above equation will be always as an odd integer.

∴ Any odd integer is of the form  6q + 1 or 6q + 3 or 6q + 5.

Hence proved.

Answer 2

QuestioN :

Use division algorithm to show that any positive odd integer is of the form of 6q+1 or 6q+3 or 6q+5, where q is some integer.​

GiveN :

• 6q+1 or 6q+3 or 6q+5,

To FiNd :

• where q is some integer.​

ANswer :

Any odd integer is of the form  6q + 1 or 6q + 3 or 6q + 5

SolutioN :

According to Euclid’s Division Lemma if we have two positive integers a and b, then there exist unique integers q and r which satisfies the condition a = bq + r where 0 ≤ r < b.

Let a be the positive odd integer which when divided by 6 gives q as quotient and r as remainder.

According to Euclid’s division lemma

a = bq + r

a = 6q + r………………….(1)

where, (0 ≤ r < 6)

So r can be either 0, 1, 2, 3, 4 and 5.

Case 1:

If r = 1, then equation (1) becomes

a = 6q + 1

The Above equation will be always as an odd integer.

Case 2:  

If r = 3, then equation (1) becomes

a = 6q + 3

The Above equation will be always as an odd integer.

Case 3:

If r = 5, then equation (1) becomes

a = 6q + 5

The above equation will be always as an odd integer.

∴ Hence, Any odd integer is of the form  6q + 1 or 6q + 3 or 6q + 5.

Hence proved.

_________________________