Q. Prove that no number of the type 4k+2 can be a perfect square.
Ans: If p is a prime factor of a perfect square, p2 must also be a factor of that perfect square. 4k+2 = 2(2k+1) 2 is a factor of 4k+2 but 2k+1 is odd and cannot have factor 2, so 4k+2 is not divisible by 4, and therefore cannot be a perfect square. Q. There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point? Ans: We need to calculate the LCM to find the answer. 18 = 2x3x3 ;12= 2x2x3 ; LCM = 36 Q. show that only one out of n, n+2 or n+4 is divisible by 3 where n is positive integer. Solution: When n = 1, exactly one of 1, 1+2, and 1+4 is divisible by 3, namely 1+2, since 3 is divisible by 3, and the other two 1 and 5 are not. Suppose for n < k, only one out of n, n+2, n+4 is divisible by 3 For n = k, we consider k, k+2, k+4. By the induction hypothesis, only one of k1, k+1 and k+3 is divisible by 3. We look at the three possible cases. Case 1: k1 is the one which is divisible by 3. Then k1 = 3m, for some positive integer m. Then add 1 to both sides of k1 = 3m: k1+1=3m+1 k = 3m+1 then k is NOT divisible by 3 Now add 3 to both sides of k1 = 3m: k1+3=3m+3 k+2 = 3m+3 k+2 = 3(m+1) then k+2 IS divisible by 3 Now add 5 to both sides of k1 = 3m: k1+5=3m+5 k+4 = 3m+5 k+4 = 3m+3 + 2 = 3(m+1)+2 then k+4 is NOT divisible by 3. So, we have proved case 1 for n = k Case 2: k+1 is the one which is divisible by 3. Then k+1 = 3m, for some positive integer m. Then add 1 to both sides of k+1 = 3m: k+11=3m1 k = 3m1 then k is NOT divisible by 3 Now add 1 to both sides of k+1 = 3m: k+1+1=3m+1 k+2 = 3m+1 then k+2 is NOT divisible by 3 Now add 3 to both sides of k+1 = 3m: k+1+3=3m+3 k+4 = 3m+3 k+4 = 3(m+1) so k+4 IS divisible by 3. So, we have proved case 2. Case 3: k+3 is the one which is divisible by 3. Then k+3 = 3m, for some positive integer m. Then add 3 to both sides of k+3 = 3m: k+33=3m3 k = 3(m1) then k IS divisible by 3 Now add 1 to both sides of k+3 = 3m: k+31=3m1 k+2 = 3m1 then k+2 is NOT divisible by 3 Now add 1 to both sides of k+3 = 3m: k+3+1=3m+1 k+4 = 3m+1 then k+4 is NOT divisible by 3. So, we have proved case 3. Q.If the H C F of 210 and 55 is expressible in the form 210 × 5 + 55y then find y Answer: Let us first find the H C F of 210 and 55. Applying Euclid division lemna on 210 and 55, we get 210 = 55 × 3 + 45 ....(1) Since the remainder 45 ≠ 0. So, again applying the Euclid division lemna on 55 and 45, we get 55 = 45 × 1 + 10 .... (2) Again, considering the divisor 45 and remainder 10 and applying division lemna, we get 45 = 4 × 10 + 5 .... (3) We now, consider the divisor 10 and remainder 5 and applying division lemna to get 10 = 5 × 2 + 0 .... (4) We observe that the remainder at this stage is zero. So, the last divisor i.e., 5 is the HCF of 210 and 55. ∴ 5 = 210 × 5 + 55y ⇒ 55y = 5  1050 = 1045 ⇒ y = 19 Q. Finds the H.C.F. of 65 and 117 and express it in the form of 65m+117n. Answer: Among 65 and 117; 2117 > 65 Since 117 > 65, we apply the division lemma to 117 and 65 to obtain 117 = 65 × 1 + 52 … Step 1 Since remainder 52 ≠ 0, we apply the division lemma to 65 and 52 to obtain 65 = 52 × 1 + 13 … Step 2 Since remainder 13 ≠ 0, we apply the division lemma to 52 and 13 to obtain 52 = 4 × 13 + 0 … Step 3 In this step the remainder is zero. Thus, the divisor i.e. 13 in this step is the H.C.F. of the given numbers The H.C.F. of 65 and 117 is 13 From Step 2: 13 = 65 – 52× 1 … Step 4 From Step 1: 52 = 117 – 65 × 1 Thus, from Step 4, it is obtained 13 = 65 – (117 – 65 × 1) ⇒13 = 65 × 2 – 117 ⇒13 = 65 × 2 + 117 × (–1) In the above relationship the H.C.F. of 65 and 117 is of the form 65m + 117 n, where m = 2 and n = –1 Q.Find all positive integral values of n for which n2+96 is perfect square. Answer: Let n2 + 96 = x2 ⇒ x2 – n2 = 96 ⇒ (x – n) (x + n) = 96 ⇒ both x and n must be odd or both even on these condition the cases are x – n = 2, x + n = 48 x – n = 4, x + n = 24 x – n = 6, x + n = 16 x – n = 8, x + n = 12 and the solution of these equations can be given as x = 25, n = 23 x = 14, n = 10 x = 11, n = 5 x = 10, n = 2 So, the required values of n are 23, 10, 5, and 2. Q. Prove that one of every three consecutive integers is divisible by 3. Ans: n,n+1,n+2 be three consecutive positive integers We know that n is of the form 3q, 3q +1, 3q + 2 So we have the following cases Case – I when n = 3q In the this case, n is divisible by 3 but n + 1 and n + 2 are not divisible by 3 Case  II When n = 3q + 1 put n = 2 = 3q +1 +2 = 3(q +1) is divisible by 3. but n and n+1 are not divisible by 3 Case – III When n = 3q +2 put n = 2 = 3q +1 +2 = 3(q +1) is divisible by 3. But n and n+1 are not divisible by 3 Hence one of n, n + 1 and n + 2 is divisible by 3 More solved Questions:
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