MCQ : Real Numbers - 1

Assume that √7 be a rational number. So, √7 = p/q where p and q are co prime.
7 = p²/q² 
7q² = p²    ...(1)
This shows that p² is divisible by 7, then p is divisible by 7, then for any positive integer c, it can be said that p = 7c, p² = 49c²
Then equation (1) can be written as,
7q² = 49c²
q² = 7c²
This gives that q is divisible by 7.
So, p and q has a common factor 7 which is a contradiction to the assumption that they are co prime.
Hence, √7 is an irrational number.

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
Ratio = LCM/HCF = 60/5 = 12/1 = 12:1