A number when divided by 61 gives 27 as quotient and 32 as remainder .Find the number
Let the number be x
divident = divisor×quotient + remainder
x = 61×27 + 32
= 1679
so, the number will be 1679
If two positive integers ‘a’ and ‘b’ are written as a = pq2 and b = p3q, where ‘p’ and ‘q’ are prime numbers, then LCM(a, b) =
a = pq2
b = p3q
LCM (a,b) = p3q2
The HCF and LCM of two numbers is 9 and 459 respectively. If one of the number is 27, then the other number is
Using the result, HCF × LCM = Product of two natural numbers ⇒ the other number = product of two no. = LCM *HCF
Let unknown no. = X
>> X * 27 = 9*459
X = 9*459/27
X = 153
So X=
So option B is correct answer.
If the HCF of 65 and 117 is expressible in the form 65m – 117, then the value of ‘m’ is
First of all find the HCF of 65 and 117,
117 = 1×65 + 52
65 = 1× 52 + 13
52 = 4 ×13 + 0
∴ HCF of 65 and 117 is 13.
65m - 117 = 13
65m = 117+13 = 130
∴m =130/65 = 2
If a is a non-zero rational and √b is irrational, then a√b is:
If possible let a√b be rational. Then a√b = p/q,
where p and q are non-zero integers, having no common factor other than 1.
Now, a√b = p/q ⇒ √b = p/aq……….(i)
But, p and aq are both rational and aq ≠ 0.
∵ p/aq is rational.
Therefore, from eq. (i), it follows that √b is rational.
The contradiction arises by assuming that a√b is rational.
Hence, a√b is irrational.
The decimal expansion of number
A number with terminal decimal expansions have the denominator in the form,
2m 5n where m & n ∈ W.
The number
Which the denominator is in the form,
with m = 2 , n = 3.
Hence, it has terminal decimal expansion.
Every positive even integer is of the form ____ for some integer ‘q’.
Let a be any positive integer and b = 2
Then by applying Euclid’s Division
Lemma, we have, a = 2q + r where 0 ⩽ r < 2 r = 0 or 1
Therefore, a = 2q or 2q+1
Therefore, it is clear that a = 2q i.e.,
a is an even integer in the form of 2q
The HCF of 867 and 255 is
As, 867=255 × 3 +102
255 = 102 × 2 + 51
102 = 51 × 2 + 0
So, HCF (867,255) = 51
If HCF(a, b) = 12 and a × b = 1800, then LCM(a, b) is
Using the result, HCF × LCM = Product of two natural numbers ⇒ LCM (a, b) = 1800/12 = 150
If d is the HCF of 56 and 72, then values of x,y satisfying d = 56 x+72y :
Since, HCF of 56 and 72, by Euclid’s divsion lemma,
72 = 56 × 1 + 16 ……….(i)
56 = 16 ×3 + 8 ……….(ii)
16 = 8× 2 + 0 ……….(iii)
∴ HCF of 56 and 72 is 8.
∴ 8 = 56 – 16× 3
8 = 56 – (72 – 56 ×1) ×3
[From eq. (i) : 16 = 72 – 56× 1]
8 = 56 – 3 ×72 + 56× 3
8 = 56 × 4 + (–3) × 72
∴ x = 4,y = −3
The number (√3+√5)2 is
Since √3 and √5 both are irrational number therefore (√3+√5)2 is an irrational number.
The largest number which divides 70 and 125, leaving remainders 5 and 8 respectively, is
Since, 5 and 8 are the remainders of 70 and 125, respectively. Thus, after subtracting these remainders from the numbers, we have the numbers 65 = (70-5),
117 = (125 – 8), which is divisible by the required number.
Now, required number = HCF of 65,117 [for the largest number]
For this, 117 = 65 × 1 + 52 [∵ dividend = divisior × quotient + remainder]
⇒ 65 = 52 × 1 + 13
⇒ 52 = 13 × 4 + 0
∴ HCF = 13
Hence, 13 is the largest number which divides 70 and 125, leaving remainders 5 and 8.
Every positive odd integer is of the form ________ where ‘q’ is some integer.
Let a be any positive integer and b = 2.
Then by applying Euclid’s Division
Lemma,
we have, a = 2q+r
where 0 ⩽ r < 2 ⇒ r = 0 or 1 ∴ a = 2q or 2q+1
Therefore, it is clear that a = 2q i.e., a is an even integer.
Also 2q and 2q+1 are two consecutive integers, therefore, 2q+1 is an odd integer.
The LCM of 23×32 and 22×33 is
L.C.M. of 23×33 and 22×32 is the product of all prime numbers with greatest power of every given number = 23×33
The HCF of the smallest prime number and the smallest composite number is
Smallest prime number = 2 and smallest composite number = 4
∴ HCF (2, 4) = 2
Which of the following is false:
H.C.F.(p,q,r)× L.C.M.(p,q,r) ≠ p×q×r. This condition is only applied on HCF and LCM of two numbers.
The number is
Since √2 and √5 both are irrational number therefore is an irrational number.
The decimal expansion of 987/10500 will terminate after
Here, in the denominator of the given fraction the highest power of prime factor 5 is 3, therefore, the decimal expansion of the rational number will terminate after 3 decimal places.
For any two positive integers a and b, there exist (unique) whole numbers q and r such that
Euclid’s Division Lemma states that for given positive integer a and b, there exist unique integers q and r satisfying a = bq+r ; 0 ⩽ r < b.
The least positive integer divisible by 20 and 24 is
Least positive integer divisible by 20 and 24 is LCM (20, 24). 20
= 22×5 24=23×3
∴ LCM (20, 24) = 23x 3 x 5 = 120
The largest number which divides 245 and 1029 leaving remainder 5 in each case is
When 245 and 1029 are divided by the required number then there is left 5 as remainder. It means that 245 - 5 = 240 and 1029 - 5 = 1024 will be completely divisible by the required number.
Now we determine the HCF of 240 and 1024 by Euclid's Algorithm.
1024 = 240 x 4 + 64
240 = 64 x 3 + 48
64 = 48 x 1 + 16
48 = 16 x 3 + 0
Since remainder comes zero with last divisor 16,
required number = 16
What is the number x? The LCM of x and 18 is 36. The HCF of x and 18 is 2.
LCM x HCF = First number x Second number
∴ Required number
If ‘a’ and ‘b’ are both positive rational numbers, then
= (a−b)
Since a and b both are positive rational numbers, therefore difference of two positive rational numbers is also rational.
The decimal expansion of 21/24 will terminate after
Here, in the denominator of the given fraction the highest power of prime factor 2 is 3, therefore, the decimal expansion of the rational numberwill terminate after 3 decimal places.
If the HCF of 65 and 117 is expressible in the form 65m – 117, then the value of m is
Find the HCF of 65 and 117,
117 = 1×65 + 52
65 = 1× 52 + 13
52 = 4 ×13 + 0
∴ HCF of 65 and 117 is 13.
65m - 117 = 13
65m = 117+13 = 130
∴m =130/65 = 2
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