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How many three digit numbers are divisible by 7? [Ex 5.2, 13 - AP]

Ex 5.2, 13 - Chapter 5 Class 10 Arithmetic Progressions - Part 2
Ex 5.2, 13 - Chapter 5 Class 10 Arithmetic Progressions - Part 3


Transcript

Ex 5.2, 13 How many three digit numbers are divisible by 7 Numbers divisible by 7 are 7, 14, 21, 28, …….. Lowest 3 digit number 100/7 = 142/7 101/7 = 143/7 102/7 = 144/7 105/7 = 15 So lowest number is 105 Highest 3 digit number, 999/7 = 1425/7 998/7 = 1424/7 997/7 = 1423/7 994/7 = 142 Highest number is 994 Thus, our series will be 105, 112, 119, 126 …. 994 Since difference is same, it is an AP Last term of this series is 994 We need to find number of terms, i.e., n Here an = 994 , a = 105 , d= 7 Putting values in formula an = a + (n – 1) d 994 = 105 + (n – 1) × 7 994 = 105 + n (7) – 1 × (7) 994 = 105 + 7n – 7 994 – 105 + 7 = 7n 896 = 7n 896/7 = n 128 = n n = 128 Therefore, 128 3-digit numbers are divisible by 7

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.