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Ex. 1. Is the number 347 prime ? Sol. Here nearest square root is 19 and prime numbers less than it are 2, 3, 5, 7, 11, 13, 17. Since 347 is not divisible by any one of these so it is a prime number. Ex. 2. Is 191 a prime number ? Sol. Since 14 is nearest square root and prime numbers less than 14 are 2, 3, 5, 7, 11, 13. Now 191 is not divisible by any one of them so it is a prime number. Ex. 3. What is the sum of first 10 natural numbers ? Sol. We know that sum of first n [n(n + 1)] natural numbers = . The first 10 2 natural number are 1 to 10. Þ Sum is [10 (10 + 1)] (10 × 11) = = = 55 2 2 Ex. 4. Find the sum of squares of first 10 natural numbers. Sol. Square of first 10 natural numbers are 1, 4, 9, 16, 25, 36, 49, 64, 81, 100. Sum of squares of first n natural numbers
Ex. 8. Simplify : 7 × [{6 (5 + 7)} - 2] Sol. 7 × [ { 6 (5 + 7) } - 2 ] = 7 × [(6 × 12) - 2] = 7 × [72 - 2] = 7 × 70 = 490. Ex. 9. What is the least number that must be added to 486 to make this number exactly divisible by 7 ? Sol. We know that Dividend = (Divisor × Quotient) + Remainder. So if we divide 486 by 7 we get 3 as a remainder. (486 = 69 × 7 + 3) Þ The number to be added = 7 - 3 = 4. Ex. 10. Find the least number of 3 digits which is exactly divisible by 17. Sol. We know that the least number of 3 digits is 100. If we divide 100 by 17 we get 15 as a remainder. i.e. 100 = 5 × 17 + 15 Þ To make it exactly divisible by 17 we should add (17 - 15) = 2 to 100. Þ Required least number is = 100 + 2 = 102 Ex. 11. Find nearest number to 182 which is exactly divisible by 6. Sol. If we divide 182 by 6 than we get 2 as a remainder.We see that 2 < 6 Þ nearest number to 182 which is exactly divisible by 6 will be = (182 - 2) = 180. Ex. 12. Find the value of 986 × 93 + 986 × 7 Sol. 986 × 93 + 986 × 7 = 986 × (93 + 7) = 986 × 100 = 98600 Ex. 13. Find the total of even numbers from 1 to 100. Sol. We know that from 1 to 100, total numbers will be 100. Þ There will be 50 even numbers. So sum of them is 50 × (50 + 1) = 2550 [because sum of first n even numbers = n (n + 1)] Ex. 14. Find the total of odd numbers from 1 to 100. Sol. We know that sum of first n odd numbers = n2. Now we have 50 odd numbers between 1 to 100. Þ Sum is (50)2 i.e. 2500
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[n(n + 1)(2n + 1)] 3 6 [10 (10 + 1)(2 × 10 + 1)] Þ S = 6 (10 × 11 × 21) when n = 10 Þ S = = 385. 6 Ex. 5. Simplify: 405 × 405 Sol. 405 × 405 = (400 + 5) (400 + 5) = (400 + 5)2 = (400)2 + (5)2 + 2 × (400) × 5 = 160000 + 25 + 4000 = 164025 Ex. 6. Simplify : 98 × 98 Sol. 98 × 98 = (100 - 2) × (100 - 2) = (100 - 2)2 = (100)2 + (2)2 - 2 × (100) × 2 = 10,000 + 4 - 400 = 9604 Ex. 7. Simplify : 102 × 98 Sol. 102 × 98 = (100 + 2) × (100 - 2) = (100)2 - (2)2 = 10000 - 4 = 9996
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Ex. 15. Find how many numbers up to 100 are divisible by 7. Sol. We see that if we divide 100 by 7 We get 100 = 14 × 7 + 2. The quotient is 14 therefore there are 14 numbers up to 100 which are divisible by 7.