MCQ : Real Numbers - 1

Lets assume that √7 is rational number. ie √7=p/q.
suppose p/q have common factor then
we divide by the common factor to get √7 = a/b were a and b are co-prime number.
that is a and b have no common factor.
√7 =a/b co- prime number
√7= a/b
a=√7b
squaring
a²=7b² .......1
a² is divisible by 7
a=7c
substituting values in 1
(7c)²=7b²
49c²=7b²
7c²=b²
b²=7c²
b² is divisible by 7
that is a and b have atleast one common factor 7. This is contradicting to the fact that a and b have no common factor.This happend because of our wrong assumption.
hence,√7 is irrational

Euclid’s division Lemma:

It tells us about the divisibility of integers. It states that any positive integer ‘a’ can be divided by any other positive integer ‘ b’ in such a way that it leaves a remainder ‘r’.

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.

Here, a= Dividend, b= Divisor, q= quotient and r = Remainder.

Given : a = 3q+r

In this question ,

b = 3

The values 'r’ can take 0 ≤ r < 3.

We know that, odd integers are 1, 3, 5, ...
So, it can be written in the form of 2q + 1,
where, q = integer = Z
or 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. Then, 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.

Greatest 5-Digit 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 5-digit number divisible by 8 and 9 = 99999 - 63 = 99936

Required number = 99936 + 5 = 99941

As the case is of 2 consecutive integers, one will be odd and the other one even, so their product will always be even which eventually will be divisible by an even no., i.e. 2, here.

Here, a = x³y² and b = xy³.
⇒ a = x _* x _* x _* y _* y and b = xy _* y _* y
∴ LCM(a, b) = x _* x _* x _* y _* y _* y = x³ _* y³ = x³y³
LCM = x³y³

If n is even then it is not possible for n=2.

If n is natural number then it is not possible for n=1,2.

If n is integer similarly n²- 1 is not divisible by 8.

If n is odd the possible for n is 3,5,7. Thus the n is odd number.

As per question, we have,
p = ab² = a × b × b
q = a³b = a × a × a × b
So, their Least Common Multiple (LCM) = a³ × b²

L.C.M. of 3, 4, 5, 6, 8 = 120 ;

Required number =2 x 2 x 2 x 2 x 3 x 3 x 5 x 5 = 3600.

5, 15 = 5 x 3, 20 = 2 x 2 x 5
LCM(5, 15, 20) = 5 x 3 x 2 x 2 = 60
HCF(5, 15, 20) = 5