(i) 225 > 135 we always divide greater number with smaller one.
Divide 225 by 135 we get 1 quotient and 90 as remainder so that
225= 135 × 1 + 90
Divide 135 by 90 we get 1 quotient and 45 as remainder so that
135= 90 × 1 + 45
Divide 90 by 45 we get 2 quotient and no remainder so we can write it as
90 = 2 × 45+ 0
As there are no remainder so divisor 45 is our HCF.
(ii) 38220 > 196 we always divide greater number with smaller one.
Divide 38220 by 196 then we get quotient 195 and no remainder so we can write it as
38220 = 196 × 195 + 0
As there is no remainder so divisor 196 is our HCF.
(iii) 867 > 255 we always divide greater number with smaller one.
Divide 867 by 255 then we get quotient 3 and remainder is 102 so we can write it as
867 = 255 × 3 + 102
Divide 255 by 102 then we get quotient 2 and remainder is 51 so we can write it as
255 = 102 × 2 + 51
Divide 102 by 51 we get quotient 2 and no remainder so we can write it as
102 = 51 × 2 + 0
As there is no remainder so divisor 51 is our HCF.
By: Admin
(i) 225 > 135 we always divide greater number with smaller one.
Divide 225 by 135 we get 1 quotient and 90 as remainder so that
225= 135 × 1 + 90
Divide 135 by 90 we get 1 quotient and 45 as remainder so that
135= 90 × 1 + 45
Divide 90 by 45 we get 2 quotient and no remainder so we can write it as
90 = 2 × 45+ 0
As there are no remainder so divisor 45 is our HCF.
(ii) 38220 > 196 we always divide greater number with smaller one.
Divide 38220 by 196 then we get quotient 195 and no remainder so we can write it as
38220 = 196 × 195 + 0
As there is no remainder so divisor 196 is our HCF.
(iii) 867 > 255 we always divide greater number with smaller one.
Divide 867 by 255 then we get quotient 3 and remainder is 102 so we can write it as
867 = 255 × 3 + 102
Divide 255 by 102 then we get quotient 2 and remainder is 51 so we can write it as
255 = 102 × 2 + 51
Divide 102 by 51 we get quotient 2 and no remainder so we can write it as
102 = 51 × 2 + 0
As there is no remainder so divisor 51 is our HCF.
By: Admin
Let take a as any positive integer and b = 6.
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 6q + 0 , 6q + 1 , 6q + 2,6q + 3, 6q + 4, 6q + 5
6q + 0
6 is divisible by 2 so it is a even number
6q + 1
6 is divisible by 2 but 1 is not divisible by 2 so it is a odd number
6q + 2
6 is divisible by 2 and 2 is also divisible by 2 so it is a even number
6q +3
6 is divisible by 2 but 3 is not divisible by 2 so it is a odd number
6q + 4
6 is divisible by 2 and 4 is also divisible by 2 it is a even number
6q + 5
6 is divisible by 2 but 5 is not divisible by 2 so it is a odd number
So odd numbers will in form of 6q + 1, or 6q + 3, or 6q + 5
By: Admin
HCF (616, 32) will give the maximum number of columns in which they can march.
We can use Euclid's algorithm to find the HCF.
616 = 32 × 19 + 8
32 = 8 × 4 + 0
The HCF (616, 32) is 8.
Therefore, they can march in 8 columns each.
By: Admin
HCF (616, 32) will give the maximum number of columns in which they can march.
We can use Euclid's algorithm to find the HCF.
616 = 32 × 19 + 8
32 = 8 × 4 + 0
The HCF (616, 32) is 8.
Therefore, they can march in 8 columns each.
By: Admin
Let a be any positive integer and b = 3.
Then a = 3q + r for some integer q ≥ 0
And r = 0, 1, 2 because 0 ≤ r < 3
Therefore, a = 3q or 3q + 1 or 3q + 2
Or,
a2 = (3q)2 or (3q + 1)2 or (3q + 2)2
a2 = (9q)2 or 9q2 + 6q + 1 or 9q2 + 12q + 4
= 3 × (3q2) or 3(3q2 + 2q) + 1 or 3(3q2 + 4q + 1) + 1
= 3k1 or 3k2 + 1 or 3k3 + 1
Where k1, k2, and k3 are some positive integers
Hence, it can be said that the square of any positive integer is either of the form 3m or 3m + 1
By: Admin
Let a be any positive integer and b = 3 a = 3q + r, where q ≥ 0 and 0 ≤ r < 3
∴ a = 3q or 3q + 1 or 3q + 2
Therefore, every number can be represented as these three forms. There are three cases.
Case 1: When a = 3q,
a3 = (3q)3 = 27q3 = 9(3q)3 = 9m,
Where m is an integer such that m = 3q3
Case 2: When a = 3q + 1, a3 = (3q +1)3 a3= 27q3 + 27q2 + 9q + 1 a3 = 9(3q3 + 3q2 + q) + 1 a3 = 9m + 1
Where m is an integer such that m = (3q3 + 3q2 + q)
Case 3: When a = 3q + 2, a3 = (3q +2)3 a3= 27q3 + 54q2 + 36q + 8 a3 = 9(3q3 + 6q2 + 4q) + 8 a3 = 9m + 8
Where m is an integer such that m = (3q3 + 6q2 + 4q)
Therefore, the cube of any positive integer is of the form 9m, 9m + 1,
or 9m + 8.
By: Admin
Let a be any positive integer and b = 3 a = 3q + r, where q ≥ 0 and 0 ≤ r < 3
∴ a = 3q or 3q + 1 or 3q + 2
Therefore, every number can be represented as these three forms. There are three cases.
Case 1: When a = 3q,
a3 = (3q)3 = 27q3 = 9(3q)3 = 9m,
Where m is an integer such that m = 3q3
Case 2: When a = 3q + 1, a3 = (3q +1)3 a3= 27q3 + 27q2 + 9q + 1 a3 = 9(3q3 + 3q2 + q) + 1 a3 = 9m + 1
Where m is an integer such that m = (3q3 + 3q2 + q)
Case 3: When a = 3q + 2, a3 = (3q +2)3 a3= 27q3 + 54q2 + 36q + 8 a3 = 9(3q3 + 6q2 + 4q) + 8 a3 = 9m + 8
Where m is an integer such that m = (3q3 + 6q2 + 4q)
Therefore, the cube of any positive integer is of the form 9m, 9m + 1,
or 9m + 8.
We have the formula that
Product of LCM and HCF = product of number
LCM × 9 = 306 × 657
Divide both side by 9 we get
LCM = (306 × 657) / 9 = 22338
By: Admin
If any digit has last digit 10 that means it is divisible by 10 and the factors of 10 = 2 × 5.
So value 6n should be divisible by 2 and 5 both 6n is divisible by 2 but not divisible by 5 So it can not end with 0.
By: Admin
7 × 11 × 13 + 13
Taking 13 common, we get
13 (7 x 11 +1 )
13(77 + 1 )
13 (78)
It is product of two numbers and both numbers are more than 1 so it is a composite number.
7 × 6 × 5 × 4 × 3 × 2 × 1 + 5
Taking 5 common, we get
5(7 × 6 × 4 × 3 × 2 × 1 +1)
5(1008 + 1)
5(1009)
It is product of two numbers and both numbers are more than 1 so it is a composite number
By: Admin
They will be meet again after LCM of both values at the starting point.
18 = 2 × 3 × 3
12 = 2 × 2 × 3
LCM = 2 × 2 × 3 × 3 = 36
Therefore, they will meet together at the starting point after 36 minutes.
By: Admin
Let take √5 as rational number
If a and b are two co prime number and b is not equal to 0.
We can write √5 = a/b
Multiply by b both side we get b√5 = a
To remove root, Squaring on both sides, we get
5b2 = a2 … (i)
Therefore, 5 divides a2 and according to theorem of rational number, for any prime number pwhich is divides a2 then it will divide a also.
That means 5 will divide a. So we can write a = 5c
Putting value of a in equation (i) we get
5b2 = (5c)2
5b2 = 25c2
Divide by 25 we get
b2/5 = c2
Similarly, we get that b will divide by 5
and we have already get that a is divide by 5
but a and b are co prime number. so it contradicts.
Hence √5 is not a rational number, it is irrational.
By: Admin
Let take that 3 + 2√5 is a rational number.
So we can write this number as
3 + 2√5 = a/b
Here a and b are two co prime number and b is not equal to 0
Subtract 3 both sides we get
2√5 = a/b – 3
2√5 = (a-3b)/b
Now divide by 2, we get
√5 = (a-3b)/2b
Here a and b are integer so (a-3b)/2b is a rational number so √5 should be a rational number But √5 is a irrational number so it contradicts.
Hence, 3 + 2√5 is a irrational number.
By: Admin
(i) Let take that 1/√2 is a rational number.
So we can write this number as
1/√2 = a/b
Here a and b are two co prime number and b is not equal to 0
Multiply by √2 both sides we get
1 = (a√2)/b
Now multiply by b b = a√2
divide by a we get b/a = √2
Here a and b are integer so b/a is a rational number so √2 should be a rational number But √2 is a irrational number so it contradicts.
Hence, 1/√2 is a irrational number
(ii) Let take that 7√5 is a rational number.
So we can write this number as
7√5 = a/b
Here a and b are two co prime number and b is not equal to 0
Divide by 7 we get
√5 = a/(7b)
Here a and b are integer so a/7b is a rational number so √5 should be a rational number but √5 is a irrational number so it contradicts.
Hence, 7√5 is a irrational number.
(iii) Let take that 6 + √2 is a rational number.
So we can write this number as
6 + √2 = a/b
Here a and b are two co prime number and b is not equal to 0
Subtract 6 both side we get
√2 = a/b – 6
√2 = (a-6b)/b
Here a and b are integer so (a-6b)/b is a rational number so √2 should be a rational number.
But √2 is a irrational number so it contradicts.
Hence, 6 + √2 is a irrational number.
By: Admin
(i) 13/3125
Factorize the denominator we get
3125 =5 × 5 × 5 × 5 × 5 = 55
So denominator is in form of 5m so it is terminating .
(ii) 17/8
Factorize the denominator we get
8 =2 × 2 × 2 = 23
So denominator is in form of 2m so it is terminating .
(iii) 64/455
Factorize the denominator we get
455 =5 × 7 × 13
There are 7 and 13 also in denominator so denominator is not in form of 2m × 5n . so it is not terminating.
(iv) 15/1600
Factorize the denominator we get
1600 =2 × 2 × 2 ×2 × 2 × 2 × 5 × 5 = 26 × 52
so denominator is in form of 2m × 5n
Hence it is terminating.
(v) 29/343
Factorize the denominator we get
343 = 7 × 7 × 7 = 73
There are 7 also in denominator so denominator is not in form of 2m × 5n
Hence it is non-terminating.
(vi) 23/(23 × 52)
Denominator is in form of 2m × 5n
Hence it is terminating.
(vii) 129/(22 × 57 × 75 )
Denominator has 7 in denominator so denominator is not in form of 2m × 5n
Hence it is none terminating.
(viii) 6/15
divide nominator and denominator both by 3 we get 2/5
Denominator is in form of 5m so it is terminating.
(ix) 35/50 divide denominator and nominator both by 5 we get 7/10
Factorize the denominator we get
10=2 × 5
So denominator is in form of 2m × 5n so it is terminating.
(x) 77/210
simplify it by dividing nominator and denominator both by 7 we get 11/30
Factorize the denominator we get
30=2 × 3 × 5
Denominator has 3 also in denominator so denominator is not in form of 2m × 5n
(i) Since this number has a terminating decimal expansion, it is a rational number of the form p/q, and q is of the form 2m × 5n.
(ii) The decimal expansion is neither terminating nor recurring. Therefore, the given number is an irrational number.
(iii) Since the decimal expansion is non-terminating recurring, the given number is a rational number of the form p/q, and q is not of the form 2m × 5n.
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