√7 is
Assume that √7 be a rational number.
i.e. √7 = p/q, where p and q are co prime.
⇒ 7 = p^{2}/q^{2}
⇒ 7q^{2 }= p^{2} ...(1)
⇒ p^{2} is divisible by 7, i.e. p is divisible by 7
⇒ For any positive integer c, it can be said that p = 7c, p^{2} = 49c^{2}
Equation (1) can be written as: 7q^{2} = 49c^{2}
⇒ q^{2} = 7c^{2}
This gives that q is divisible by 7.
Since p and q have a common factor 7 which is a contradiction to the assumption that they are coprime.
Therefore, √7 is an irrational number.
For positive integers a and 3, there exist unique integers q and r such that a = 3q + r, where r must satisfy:
Euclid's Division Lemma states that for any two positive integers ‘a’ and ‘b’ there exist two unique whole numbers ‘q’ and ‘r’ such that, a = bq + r, where 0≤ r < b.
Where, a= Dividend, b= Divisor, q= quotient and r = Remainder.
Given: a = 3q+r
⇒ The values 'r’ can take 0 ≤ r < 3.
For some integer q, every odd integer is of the form
We know those odd integers are 1, 3, 5, ...
So, it can be written in the form of 2q + 1, where, q = integer = Z
⇒ q = ..., 1, 0,1,2,3, ...
2q + 1 = ..., 3, 1, 1, 3,5, ...
Alternate Method
Let 'a' be given positive integer. On dividing 'a' by 2, let q be the quotient and r be the remainder.
By Euclid's division algorithm, we have
a = 2q + r, where 0 ≤ r < 2
⇒ a = 2q + r, where r = 0 or r = 1
⇒ a = 2q or 2q + 1
when a = 2q + 1 for some integer q, then clearly a is odd.
Find the greatest number of 5 digits, that will give us remainder of 5, when divided by 8 and 9 respectively.
Greatest 5Digit number = 99999
LCM of 8 and 9,
8 = 2 × 2 × 2
9 = 3 × 3
LCM = 2 × 2 × 2 × 3 × 3 = 72
Now, dividing 99999 by 72, we get
Quotient = 1388
Remainder = 63
So, the greatest 5digit number divisible by 8 and 9 = 99999  63 = 99936
Required number = 99936 + 5 = 99941
The product of two consecutive integers is divisible by
As the case is of 2 consecutive integers, one will be odd and the other one even,
⇒ Their product will always be even i.e. it will always be divisible by an even no., i.e. 2, here.
If two positive integers a and b are written as a = x^{3}y^{2} and b = xy^{3}, where x, y are prime numbers, then LCM(a, b) is
Here, a = x^{3}y^{2} and b = xy^{3}.
⇒ a = x _{*} x _{*} x _{*} y _{*} y and b = xy _{*} y _{*} y
∴ LCM(a, b) = x _{* }x _{* }x _{* }y _{*} y _{*} y = x^{3} _{*} y^{3} = x^{3}y^{3}
LCM = x^{3}y^{3}
n^{2}  1 is divisible by 8 if n is
Option A: If n is integer similarly n² 1 is not divisible by 8.
Option B: If n is a natural number then it is not possible for n=1,2.
Option C: If n is odd the possible for n is 3,5,7...
Option D: If n is even then it is not possible for n=2.
Thus, the option C is correct.
If two positive integers p and q can be expressed as p = ab^{2} and q = a^{3}b; where a, b being prime numbers, then LCM (p, q) is equal to
As per question, we have,
p = ab^{2} = a × b × b
q = a^{3}b = a × a × a × b
So, their Least Common Multiple (LCM) = a^{3} × b^{2}
The least perfect square number which is divisible by 3, 4, 5, 6 and 8 is
L.C.M. of 3, 4, 5, 6, 8 = 2 × 2 × 2 × 3 × 5 = 120
Pair of 2, 3 and 5 is not completed.
To make it a perfect square, the number should be multiplied by 2, 3, 5.
Required number = 120 x 2 x 3 x 5 = 3600.
The ratio between the LCM and HCF of 5,15, 20 is:
Factors are following:
5 = 5 x 1
15 = 5 x 3
20 = 2 x 2 x 5
LCM = 5 x 3 x 2 x 2 = 60
HCF = 5
Ratio = LCM/HCF = 60/5 = 12/1 = 12:1
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