(i) Here, we have to find H.C.F of 135 and 225
First divide divide the larger integer smaller integer
Since, 225 > 135
Therefore, by Euclid’s Division algorithm
225 = 135 × 1 + 90 (i)
Here 90 ≠ 0, so proceed the same procedure further
Again by E.D.L, (E.D.L = Euclid’s division algorithm)
135 = 90 × 1 + 45 (ii)
As we know, 45 ≠ 0 therefore, again by E.D.L
90 = 45 × 2 + 0 (iii)
Here, r = 0 so we cannot proceed further. The divisor at this Stage is 45.
From (i), (ii) and (iii)
H.C.F (225, 135) = H.C.F (135, 90) = H.C.F (90, 45) = 45.
(ii) Here, we have to find H.C.F of 38220 and 196
First divide the larger integer smaller integer
Since, 3822 > 196
Therefore by Euclid’s Division Algorithm
38220 = 196 × 195 + 0
Here, r = 0 so we cannot proceed further. The divisor at this Stage is 196.
Hence, H.C.F (38220, 196) = 196.
(iii) Here, we have to find H.C.F of 867 and 255
First divide the larger integer smaller integer
Since, 867 > 255
Therefore, by Euclid’s Division algorithm
867 = 255 × 3 + 102 (i)
Remainder 102 ≠ 0, so proceed the same procedure further using E.D.L
255 = 102 × 2 + 51 (ii)
Here, 51 ≠ 0 again using E.D.L = 51 × 2
102 = 51 × 2 + 0 (iii)
Here, r = 0 so we cannot proceed further. The divisor at this Stage is 51.
From (i), (ii) and (iii)
H.C.F (867, 255) = H.C.F (255, 102) = H.C.F (102, 51) = 51.
Let a be any positive integer b = 6. Then by Euclid’s algorithm,
a = 6q + r for some integer q ≥ 0
So, values of r we get, r = 0, 1, 2, 3, 4, 5
When, r = 0, then, a = 6q + 0,
similarly for r = 1, 2, 3, 4, 5 the value of a is 6q+1, 6q+2, 6q+3, 6q+4 and 6q+5 respectively.
If a = 6q, 6q+2, 6q+4, then a is even and divisible by 2.
Therefore, any positive odd integer is in the form of 6q+1, 6q+3, 6q+5
Where q is some integer.
Total no. of army contingent members = 616
No. of army band members = 32
To find max. numbers of the same columns in which the both groups march. We have to find it.
Their highest common factor. since, 616 > 32 then, by Euclid’s algorithm
616 = 32 × 19 + 8 (i)
Here 8 ≠ 0 then by again using Euclid algorithm
32 = 8 × 4 + 0 (ii)
Here, r =0 so we cannot proceed further. The divisor at this Stage is 8.
So the no. of columns is 8.
Let x be any positive integer and b = 3.
Then, by euclid’s algorithm
x = 3q + r, where q ≥ 0 and r = 0, 1, 2 [0 ≤ r ≤ b]
Case (i) : For r = 0, x = 3q, = x2 = 9q2, taking 3 as common,
x2 = 9q2 = 3 (3q2), which is of the form 3m, where m = 3q2.
Case (ii) : For r = 1, x = 3q + 1
x2 = 9q2 + 1 + 6q, taking 3 as common,
= 3 (3q2 + 2q) + 1, which is of the form 3m + 1, where m = 3q2 + 2q
Case (iii) : For r = 2, 3q + 2
x2 = 9q2 + 4 + 12q = (9q2 + 12q + 3) + 1, taking 3 as common,
= 3 (3q2 + 4q + 1) + 1, which is of the form 3m +1, where m = 3q2 + 4q + 1
Hence, x2 is either of the form 3m, 3m + 1 for some integer m.
Let a be any positive integer and b = 3. Then, by euclid’s algorithm
a = 9q + r, where q ≥ 0 and r = 0, 1, 2 [ 0 ≤ r ≤ b ]
For r = 0, x = 3q, or
For r = 1, x = 3q +1
For r = 2, x = 3q + 2
Now by taking the cube of all the three above terms, we get,
Case (i) : when r = 0, then,
x3 = (3q)3 = 27q3 = 9 (3q3) = 9m; where m = 3q
Case (ii) : when r = 1, then,
x3 = (3q +1)3 = (3q)3 + 13 +3 × 3q × 1 (3q + 2 ) = 27q3 + 1 +27q2 + 9q
Taking 9 as common factor, we get,
x3 = 9 (3q3 +3q2 + q) +1
Putting (3q3 + 3q2 + q) = m, we get,
x3 = 9m + 1
Case (iii) : when r = 2, then,
x3 = (3q + 2)3 = (3q)3 + 23 + 3 × 3q × 2 (3q + 2) = 27q3 + 54q2 + 36q + 8
Taking 9 as common factor , we get
x = 9 (3q + 6q + 4q) + 8
Putting (3q + 6q + 4q) = m, we get,
x = 9m + 8,
Therefore, it is proved that the cube of any positive integer is of the form 9m, 9m + 1, 9m + 8.
(i) 140
By taking the L.C.M of 140, we will get the product of its prime factors.
Therefore, 140 = 2 × 2 × 5 × 7 ×1 = 22 × 5 × 7
(ii) 156
By taking the L.C.M of 156, we will get the product of its prime factors.
Therefore, 156 = 2 × 2 × 13 × 3 × 1 = 22 × 13 × 3
(iii) 3825
By taking the L.C.M of 3825, we will get the product of its prime factors.
Therefore, 3825 = 3 × 3 × 5 × 5 × 17 × 1 = 32 × 52 × 17 × 1
(iv) 5005
By taking the L.C.M of 5005, we will get the product of its prime factors.
5005 = 5 × 5 × 11 × 13 × 1 = 5 × 7 × 11 × 1
(v) 7429
By taking the L.C.M of 7429, we will get the product of its prime factors.
Therefore, 7429 = 17 × 19 × 13 × 1
(i) 26 and 91
First we have to find the L.C.M and H.C.F of 26, 91 using fundamental theorem of arithmetic,
26 = 13 × 2
91 = 13 × 7
For L.C.M - list all the prime factors (only once) of 26, 91 with their greatest power.
L.C.M (26, 91) = 13 × 2 × 7 = 182
For H.C.F – write all the common factors (only once) with their smallest exponent.
H.C.F (26, 91) = 13
Verification : L.C.M (26, 91) × H.C.F (26, 91) = 26 × 91
182 × 13 = 2366, => 2366 = 2366
Hence, proved.
(ii) 510 and 92
First we have to find the L.C.M and H.C.F of 510, 92 using fundamental theorem of arithmetic,
510 = 2 × 3 × 5 × 17
92 = 2 × 2 × 23
For L.C.M - list all the prime factors (only once) of 510, 92 with their greatest power.
L.C.M (510, 92) = 2 × 2 × 3 × 5 × 17 × 23 = 23460
For H.C.F – write all the common factors (only once) with their smallest exponent.
H.C.F (510, 92) = 2
Verification : L.C.M (510, 92) × H.C.F (510, 92) = 510 × 92
23460 × 2 = 46920, => 46920 = 46920
Hence, proved.
(iii) 336 and 54
First we have to find the L.C.M and H.C.F of 336, 54 using fundamental theorem of arithmetic,
336 = 2 × 2 × 2 × 2 × 3 × 7 = 24 × 3 × 7
54 = 2 × 3 × 3 × 3 = 2 × 33
For L.C.M - list all the prime factors (only once) of 336, 54 with their greatest power.
L.C.M (336, 54) = 3024
For H.C.F – write all the common factors (only once) with their smallest exponent.
H.C.F (336, 54) = 2 × 3 = 6
Verification : L.C.M (336, 54) × H.C.F (336, 54) = 336 × 54
3024 × 6 = 18144 => 18144 = 18144
Hence, proved.
(i) 12, 15 and 21
12 = 2 × 3 × 2
15 = 3 × 5
21 = 3 × 7
L.C.M (12, 15, 21) = 2 × 3 × 2 × 5 × 7 = 420
H.C.F (12, 15, 21) = 3
(ii) 17, 23 and 29
17 = 17 × 1
23 = 23 × 1
29 = 29 ×1
L.C.M (17, 23, 29) = 11339
H.C.F (17, 23, 29) = 1
(iii) 8, 9 and 25
8 = 2 × 2 × 2 × 1
9 = 3 × 3 × 1
25 = 5 × 5 × 1
L.C.M (8, 9, 25) = 1800
H.C.F (8, 9, 25) = 1
Given,
H.C.F (306, 657) = 9
First number (a) = 306
Second number (b) = 657
As we know, H.C.F × L.C.M = a × b
9 × L.C.M = 306 × 657
L.C.M = (306 × 657) / 9 = 22338
Hence, L.C.M (306, 657) = 22338
Suppose the number 6n for any n € N ends with zero. In this case 6n is divisible by 5.
Prime factorization of 6n = (2 × 3)n
Therefore, the prime factorization of 6n doesn’t contain prime number 5.
Hence, it is clear that for any natural number n, 6n is not divisible by 5 and thus it proves that 6n cannot end with the digit 0 for any natural number n.
Composite numbers are factors other than 1 and itself.
From here we can observe that :
(a) 7 × 11 × 13 + 13 = 13 (7 × 11 × 1 + 1) = 13 (77 + 1)
= 13 × 78
= 13 × 13 × 6
Therefore, this expression has 6 and 13 as its factors. Hence, it is a composite number.
(b) 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 = 5 (7 × 6 × 1 × 4 × 3 × 2 × 1 + 1)
= 5 × (1008 + 1)
= 5 × 1009
1009 can not be factored further. Therefore, this expression has 5 and 1009 as its factors.
Hence, it is a composite number.
Time taken by Sonia to drive one round = 18 min
Time taken by Ravi to drive one round = 15 min
Let they both meet after time ‘t’. Here, ‘t’ is L.C.M of 18, 12
Now, 18 = 2 × 3 × 3
12 = 2 × 3 × 2
L.C.M (18, 12) = 22 × 32 = 36
Thus, they both meet after 36 minutes at the starting point.
Note: - If x be a rational number, then x can be expressed in the form p/q where p and q are Co- primes. Then, if the prime factorization of denominator (q) is in the form of 2m × 5n where n, m are non- negative integers, then x has a decimal expansion which terminates. If the prime factorization is not in this form of 2m × 5n then x has a decimal expansion which is non terminating.
(i) 13 / 3125
= 5 × 5 × 5 × 5 × 5 = 5⁵ (on factorization of q)
Since, its factorization contains only power of 5.
Therefore, it has a terminating decimal expansion.
(ii) 17 / 5
= 2 × 2 × 2 = 2³ (on factorization of q)
Since, its factorization contains only power of 2.
Therefore, it has a terminating decimal expansion.
(iii) 64 / 455
= 5 × 7 × 13 = 5¹ × 7 × 13 (on factorization of q)
Since the factorization of not in the 2m ×5n
Therefore, it has a non -terminating decimal expansion.
(iv) 15 / 1600
= 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5 = 2⁶ × 5² (on factorization of q)
Since the factorization of q is in the form 2m ×5n
Therefore, it has a terminating decimal expansion
(V) 29 / 343
= 7 × 7 × 7 = 7³ (on factorization of q)
Since the factorization of denominator is not in the form 2m × 5n
Therefore it has a non -terminating repeating decimal
(vi) 23 / 23 52
Since factorization of q already given and it is in the form 2m × 5n
Therefore it has a terminating decimal expansion.
(vii) 129 / 22 57 75
Since factorization of q already given but it is not in the form 2m × 5n
Therefore it has a terminating decimal expansion.
(viii) 6 / 15 = 2/5
5 = 5 × 1 (on factorization of q)
Since 5 is the only factor in denominator.
Therefore it has a non -terminating repeating decimal expansion.
(ix). 35 / 50 = 7 / 10
10 = 2 × 5 = 2¹ × 5 (on factorization of q)
Since the factorization of denominator is in the form 2m × 5n
Therefore it has a terminating decimal expansion.
(x) 77 / 210 = 11/30
30 = 2 × 3 × 5 × 7 = 2¹ × 3 × 5 × 7. (on factorization of q)
Since the factorization of denominator is not in the form 2m × 5n
Therefore it has a non -terminating repeating decimal expansion.
(i) 13 / 3125
(ii) 17 / 8
(iii) 64/455 is non terminating repeating decimal expansion.
(iv) 15 / 1600
(v) 29/343 is non terminating repeating decimal expansion.
(vi) 23 / 23 52
(vii) 120 / 23 57 75 is non terminating repeating decimal expansion.
(viii) 6/15 = 2/5
Q:
Note: A rational number has a decimal expansion which is either terminating or non -Terminating repeating otherwise the given number is irrational.
Since, the given decimal expansion is terminating. Therefore it is a rational number and we know that the prime factorization of denominator of rational 2 and 5 or both.
Since, the given decimal expansion is non- terminating non- repeating. Therefore it is irrational number.
Since, the given decimal expansion is non- terminating repeating. Therefore it is a rational number and we know that the prime factorization of denominator of rational 2 and 5 or both.
Let us consider √5 is a rational number.
Therefore, √5 = pq, where p and q are integers and q ≠ 0
If pa and q have any common factor ,then dividing by that common factor,
We have, √5 = ab, where a and b are co primes.
a = √5b
On squaring, a² = 5b² ........... (1)
Due to the presence of 5 on RHS, we say that 5 is a factor of a².
5 divides a²
Since, a is prime , therefore 5 divides a [ theorem ]
Therefore , a = 5k , where k is an integer.
Putting a = 5k in (1) , we have
25k² = 5b²
b² = 5k²
This shows that 5divides b². But b is a prime no and so 5 divides b also.
Thus 5 is a common factor of a and b . This is contradiction to the fact that
a and b have no common factor other than one.
Thus our consideration is wrong and so √ 5 is not a rational number.
Hence, proved.
Let us assume 3 + 2√5 is a rational number.
Therefore, 3 + 2√5 = p/q where p and q are co primes and q ≠ 0.
3 + 2√5 = ab
On solving, 2√5 =(a/b) - 3
√5 =1/2 (a/b - 3)
Since a, b are integers and 1/2 (a/b-3 ) is also a rational number.
But we know √5 is an irrational number.
Thus our assumption is wrong. 3 + 2√5 is not a rational number.
Hence proved.
(i) Let us assume 1/√2 is a rational number.
1/√2 = p/q , where q ≠ 0 and p and q are co primes.
On reciprocal,
√2 = qp ................(1)
Since, q and p are integers and q/p is also a rational number
As we know √2 is an irrational number.
From (1)
√2 ≠ q/p
Thus our assumption is wrong 1/√2 is not a rational number.
Hence, proved
(ii) Let us suppose 7√5 is a rational number.
7√5 = p/q, where p and q are co primes and q ≠ 0
On solving , √5 = (p/q)7 .....................(1)
Since p, q and 7 integers and (p/q)7 is also a rational number.
And we know √5 is an irrational number.
From (1)
√5 ≠ (p/q) / 7
So our supposition Is wrong 7√5 is not a rational number.
Hence, proved.
(iii) Let us suppose 6 + √2 is a rational number.
6 + √2 = a/b, where a, b are co primes and b ≠ 0.
On solving,
√2 = a/b - 6 .....................(1)
Since a, b and 6 are integers and a/b - 6 is also a rational number.
And we know that √2 is an irrational number.
From (1)
√2 ≠ a/b - 6
Thus our Superposition is wrong 6√2 is not a rational number.
Hence, proved.
