Q.1. Based on Euclid’s algorithm: a = bq + r ; Using Euclid’s algorithm: Find the HCF of 825 and 175.
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 65m + 117 n, where m = 2 and n = –1
1oth maths term-1
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 65m + 117 n, where m = 2 and n = –1
1oth maths term-1
