Ans:Since 825>175, we apply division lemma to 825 and 175 to get825 = 175 x 4 + 125.

Since r ≠ 0, we apply division lemma to 175 and 125 to get

175 = 125 x 1 + 50

Again applying division lemma to 125 and 50 we get,

125 = 50 x 2 + 25.

Once again applying division lemma to 50 and 25 we get.

50 = 25 x 2 + 0.

Since remainder has now become 0, this implies that HCF of 825 and 125 is 25.

Q.2. Based on Showing that every positive integer is either of the given forms:

Solved example:

Prove that every odd positive integer is either of the form 4q + 1 or 4q + 3 for some integer q.

Ans: Let a be any odd positive integer (first line of problem) and let b = 4. Using division Lemma we can write a = bq + r, for some integer q, where 0≤r<4. So a can be 4q, 4q + 1, 4q + 2 or 4q + 3. But since a is odd, a cannot be 4q or 4q + 2. Therefore any odd integer is of the form 4q + 1 or 4q + 3.

Q.Find H C F (26,91) if LCM(26,91) is 182

Sol: We know that LCM x HCF = Product of numbers.

or 182 x HCF = 26 x 91

or HCF = 26 x 91 = 13

Hence HCF (26, 91) = 13.

Q. prove that √5 is irrational.

Solution: let us assume on the contrary that √5 is rational. That is we can find co-primes a and b b (≠0) such that √5 = a/b.

Or √5b = a.

Squaring both sides we get

5b2 = a2.

This means 5 divides a2. Hence it follows that 5 divides a.

So we can write a = 5c for some integer c.

Putting this value of a we get

5b2 = (5c)2

Or 5b2 = 25c2

Or b2 = 5b2.

It follows that 5 divides b2. Hence 5 divides b.

Now a and b have at least 5 as a common factor.

But this contradicts the fact that a and b are co-primes.

This contradiction has arisen because of our incorrect assumption that √5 is rational. Hence it follows that √5 is irrational.

Q. prove that product of three consecutive positive integers is divisible by 6?

Ans: Let three consecutive positive integers be, n, n + 1 and n + 2.

Whenever a number is divided by 3, the remainder obtained is either 0 or 1 or 2.

∴ n = 3p or 3p + 1 or 3p + 2, where p is some integer.

If n = 3p, then n is divisible by 3.

If n = 3p + 1, then n + 2 = 3p + 1 + 2 = 3p + 3 = 3(p + 1) is divisible by 3.

If n = 3p + 2, then n + 1 = 3p + 2 + 1 = 3p + 3 = 3(p + 1) is divisible by 3.

So, we can say that one of the numbers among n, n + 1 and n + 2 is always divisible by 3.

⇒ n (n + 1) (n + 2) is divisible by 3.

Similarly, whenever a number is divided 2, the remainder obtained is 0 or 1.

∴ n = 2q or 2q + 1, where q is some integer.

If n = 2q, then n and n + 2 = 2q + 2 = 2(q + 1) are divisible by 2.

If n = 2q + 1, then n + 1 = 2q + 1 + 1 = 2q + 2 = 2 (q + 1) is divisible by 2.

So, we can say that one of the numbers among n, n + 1 and n + 2 is always divisible by 2.

⇒ n (n + 1) (n + 2) is divisible by 2.

Since, n (n + 1) (n + 2) is divisible by 2 and 3. ∴ n (n + 1) (n + 2) is divisible by 6.

Q. Express HCF of F 65 and 117 in the form of 65m +117n

Ans: Among 65 and 117; 117 > 65

Since 117 > 65, we apply the division lemma to 117 and 65 to obtain

117 = 65 x 1 + 52

**… Step 1**

Since remainder 52 ≠ 0, we apply the division lemma to 65 and 52 to obtain

65 = 52 x 1 + 13

**… Step 2**

Since remainder 13 ≠ 0, we apply the division lemma to 52 and 13 to obtain

52 = 4 x 13 + 0

**… Step 3**

In this step the remainder is zero. Thus, the divisor i.e. 13 in this step is the H.C.F. of the given numbers

The H.C.F. of 65 and 117 is 13

From

**Step 2**:

13 = 65 – 52 x 1

**… Step 4**

From

**Step 1**:

52 = 117 – 65 x 1

Thus, from

**Step 4**, it is obtained

13 = 65 – (117 – 65 x 1) x 1

⇒13 = 65 x 2 – 117

⇒13 = 65 x 2 + 117 x (–1)

In the above relationship the H.C.F. of 65 and 117 is of the form 65

*m*+ 117

*n*, where

*m*= 2 and

*n*= –1

**1oth maths term-1**