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Let us assume  is rational. It means  can be written in the form  where p and q are co-primes and  Squaring both sides we obtain From the above equation, we can see that p2 is divisible by 5, Therefore p will also be divisible by 5 as 5 is a prime number.  Therefore p can be written as 5r p = 5r  p2 = (5r)2 5q2 = 25r2 q2 = 5r2 From the above equation, we can see that q2 is divisible by 5,...
The time after which they meet again at the starting point will be equal to the LCM of the times they individually take to complete one round. Time taken by Sonia = 18 = 2 x 32 Time taken by Ravi = 12 = 22 x 3  LCM(18,12) = 22 x 32 = 36 Therefore they would again meet at the starting point after 36 minutes.
7 x 11 x 13 + 13 = (7 x 11 + 1) x 13 = 78 x 13 = 2 x 3 x 132 7 x 6 x 5 x 4 x 3 x 2 x 1 + 5 = (7 x 6 x 4 x 3 x 2 x 1 + 1) x 5 = 5 x 1008  After Solving we observed that both the number are even numbers and the number rule says that we can take atleast two common out of two numbers. So that the number is composite number.
By prime factorizing we have 6n = 2n x 3n A number will end with 0 if it has at least 1 as the power of both 2 and 5 in its prime factorization. Since the power of 5 is 0 in the prime factorization of 6n we can conclude that for no value of n 6n will end with the digit 0.
As we know the product of HCF and LCM of two numbers is equal to the product of the two numbers we have HCF (306, 657) x LCM (306, 657) = 306 x 657
The given numbers are written as the product of their prime factors as follows 8 = 23 9 = 32 25 = 52 HCF = 1 LCM = 23 x 32 x 52 = 1800
The given numbers are written as the product of their prime factors as follows 17 = 1 x 17 23 = 1 x 23 29 = 1 x 29 HCF = 1 LCM = 17 x 23 x 29 = 11339
The numbers can be written as the product of their prime factors as follows 12 = 22 x 3 15 = 3 x 5 21 = 3 x 7 HCF = 3  LCM = 22 x 3 x 5 x 7 = 420
336 is expressed as the product of its prime factor as 336 = 24 x 3 x 7 54 is expressed as the product of its prime factor as 54 = 2 x 33 HCF(336,54) = 2 x 3 = 6 LCM(336,54) = 24 x 33 x 7 = 3024 HCF x LCM = 6 x 3024 = 18144 336 x 54 = 18144 336 x 54 = HCF x LCM Hence Verified
The number can be expressed as the product of prime factors as 510 = 2 x 3 x 5 x 17 92 = 22 x 23 HCF(510,92) = 2 LCM(510,92) = 22 x 3 x 5 x 17 x 23 = 23460 HCF x LCM = 2 x 23460 = 46920 510 x 92 = 46920 510 x 92 = HCF x LCM Hence Verified
26 = 2 x 13 91 = 7 x 13 HCF(26,91) = 13 LCM(26,91) = 2 x 7 x 13 = 182 HCF x LCM = 13 x 182 = 2366 26 x 91 = 2366 26 x 91 = HCF x LCM Hence Verified
The given number can be expressed as the product of their prime factors as follows
The given number can be expressed as the product of its prime factors as follows.
The number is expressed as the product of the prime factors as follows
The given number can be expressed as follows
The number can be as a product of its prime factors as follows
Let x be any positive integer. It can be written in the form 3q + r where  and r = 0, 1 or 2 Case 1: For r = 0 we have x3 = (3q)3 x3 = 27q3 x3 = 9(3q3) x3 = 9m Case 2: For r = 1 we have x3 = (3q+1)3 x3 = 27q3 + 27q2 + 9q + 1 x3 = 9(3q3 + 3q2 +q) + 1 x3 = 3m + 1 Case 3: For r = 2 we have x3 = (3q+2)3 x3 = 27q3 + 54q2 + 36q + 8 x3 = 9(3q3 + 6q2 +4q) + 8 x3 = 3m + 8 Hence proved.
Let x be any positive integer. It can be written in the form 3q + r where  and r = 0, 1 or 2 Case 1: For r = 0 we have x2 = (3q)2 x2 = 9q2 x2 = 3(3q2) x2 = 3m Case 2: For r = 1 we have x2 = (3q+1)2 x2 = 9q2 + 6q +1 x2 = 3(3q2 + 2q) + 1 x2 = 3m + 1 Case 3: For r = 2 we have x2 = (3q+2)2 x2 = 9q2 + 12q +4 x2 = 3(3q2 + 4q + 1) + 1 x2 = 3m + 1 Hence proved.
The maximum number of columns in which they can march = HCF (32, 616) Since 616 > 32, applying Euclid's Division Algorithm we have Since remainder  0 we again apply Euclid's Division Algorithm Since 32 > 8  Since remainder  = 0 we conclude, 8 is the HCF of  616 and  32.  The maximum number of columns in which they can march is 8.
Let p be any positive integer. It can be expressed as p = 6q + r where  and but for r = 0, 2 or 4 p will be an even number therefore all odd positive integers can be written in the form 6q + 1, 6q + 3 or 6q + 5.
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