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Q1. Find the greatest number of four digits which is divisible by 5,12 and 36.

Solution

The greatest number of four digits is 9999. Any number which is divisible by all of 5, 12 and 36 has to be a least common multiple of 5, 12 and 36.... So, L.C.M. (5, 12, 36) =2 x 2 x 3 x 3 x 5=180...Now, 9999 is not divisible by 180 So the required number has to be less than 9999.On dividing 9999 by 180 we get Quotient=55 and Remainder=99.So the required number is 9999-99=9900.  So the greatest number of four digits which is divisible by 5,12, 36 is 9900

Q2. An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?

Solution

Let x be the maximum number of columns is which the two groups can march, then x is HCF of 616 and 32. Here 616 > 32 By Euclid division algorithm, 616 = 32 × 19 + 8 32   = 8 × 4 + 0 Thus, HCF (616, 32) = 8 Hence the maximum number of columns in which they can march is 8.

Q3. Show that the square of any positive integer cannot be of the form 5q + 2 or 5q + 3 for any integer q.

Solution

Let a be any positive integer. By Euclid's division lemma, a = bm + r where b = 5 ⇒ a = 5m + r So, r can be any of 0, 1, 2, 3, 4 ∴  a = 5m when r = 0,    a = 5m + 1 when r = 1,    a = 5m + 2 when r = 2,    a = 5m + 3 when r = 3,    a = 5m + 4 when r = 4 Case I : a = 5m ⇒ a² = (5m)² = 25m² ⇒ a² = 5(5m²) = 5q, where q = 5m² Case II : a = 5m + 1 ⇒ a² = (5m + 1)² = 25m² + 10 m + 1 ⇒ a² = 5 (5m² + 2m) + 1 = 5q + 1, where q = 5m² + 2m Case III : a = 5m + 2 ⇒ a² = (5m + 2)² ⇒ a² = 25m² + 20m + 4 ⇒ a² = 5 (5m² + 4m) + 4 ⇒ a² = 5q + 4 where q = 5m² + 4m Case IV: a = 5m + 3 ⇒ a² = (5m + 3)² = 25m² + 30m + 9 ⇒ a² = 5 (5m² + 6m + 1) + 4 ⇒ a² = 5q + 4 where q = 5m² + 6m + 1 Case V: a = 5m + 4 ⇒ a² = (5m + 4)² = 25m² + 40m + 16 ⇒ a² = 5 (5m² + 8m + 3) + 1 ⇒ a² = 5q + 1 where q = 5m² + 8m + 3 From all these cases, it is clear that square of any positive integer cannot be of the form 5q + 2 or 5q + 3.

Q4. If x(≠ o) is a rational number and y is irrational number, then what can you say about the following:            (i)  x + y             (ii)  x – y             (iii) xy             (iv) x ÷ y

Solution

(i) irrational       (ii) irrational       (iii) irrational   (iv) irrational

Q5. Express 264 as a product of its prime factors:

Solution

264 = 2³ ×3×11