Question

26. Show that any positive odd integer can be written in the form 6 m+1,6 m+3 or 6
m+5 where m is a positive integer.​

Answer 1

Let take a as any positive integer and b = 6. a > b

Then using Euclid’s algorithm, we get a = 6q + r here r is remainder and value of q is more than or equal to 0 and r = 0, 1, 2, 3, 4, 5 because 0 ≤ r < b and the value of b is 6

So total possible forms will be 6q + 0, 6q + 1 , 6q + 2,6q + 3,6q + 4,6q + 5.

6q + 0 → 6 is divisible by 2 so it is an even number.

6q + 1 → 6 is divisible by 2 but 1 is not divisible by 2 so it is an odd number.

6q + 2 → 6 is divisible by 2 and 2 is also divisible by 2 so it is an even number.

6q + 3 → 6 is divisible by 2 but 3 is not divisible by 2 so it is an odd number.

6q + 4 → 6 is divisible by 2 and 4 is also divisible by 2 it is an even number.

6q + 5 → 6 is divisible by 2 but 5 is not divisible by 2 so it is an odd number.

So odd numbers will in form of 6q + 1, or 6q + 3, or 6q

Answer 2

Step-by-step explanation -

Let a be any positive integer and b = 6, where a > b

Using Euclid's algorithm :

Divident = Divisor × Quotient + Remainder

→ a = bq + r

Here -

• r = remainder
• q = 0, 1, 2, 3, 4, 5 (0 ≤ r ≤ b)
• b = 6

So, we can also write it like : a = bm + r

So, the possible form are :

→ 6q + 0, 6q + 1, 6q + 2, 6q + 3, 6q + 4, 6q + 5.

• 6q + 0 is divisible by 2. As, it is an even number.

• 6q + 1 is not divisible by 2. As, it is not an even number. It's an odd number.

• 6q + 2 is divisible by 2. As, it is an even number.

• 6q + 3 is not divisible by 2. As, it is an odd number.

• 6q + 4 is divisible by 2. As, it is an even number.

• 6q + 5 is not divisible by 2. As, it is an odd number.

•°• So, positive odd integers is in the form : 6m + 1, 6m + 3 and 6m + 5, where m is any integer.

$\rule{200}2$