    # A least number of 4 digit when increased by 5

 Solution: We will be using the concept of LCM(Least Common Multiple) to solve this. We know that the smallest 4-digit number is 1000. Hence,the LCM of 18, 24 and 32 is calculated as shown below, Therefore, LCM of 18, 24 and 32 = 2 × 2 × 2 × 2 × 2 × 3 × 3 = 288 Thus, we have 288 as the smallest number, which is exactly divisible by 18, 24, and 32. Since it's not a 4-digit number, we need to find the multiple of 288, close to 1000. Here we have 136 as the remainder. Therefore, we need to subtract 136 and add 288 to make the smallest 4 digit number exactly divisible by18, 24, and 32. So, the multiple of 288 just above 1000 is: 1000 – 136 + 288 = 1152. Hence, the smallest 4-digit number which is divisible by 18, 24, and 32 is 1152. You can also use the LCM Calculator to solve this. NCERT Solutions for Class 6 Maths Chapter 3 Exercise 3.7 Question 9 Summary: The smallest 4-digit number which is exactly divisible by 18,24, and 32, is 1152. ☛ Related Questions: India's Super Teachers for all govt. exams Under One Roof Enroll For Free Now India’s #1 Learning Platform Start Complete Exam Preparation Daily Live MasterClasses Practice Question Bank Mock Tests & Quizzes Get Started for Free Download App Trusted by 3.4 Crore+ Students 52. A least number of 4 digits when increased by 5 is completely divisible by 12, 15, 20 and 35 , the number is: / 4 अंकों की एक न्यूनतम संख्या मे 5 की वृद्धि करने पर वह 12, 15, 20 और 35 से पूरी तरह से विभाज्य है। वह संख्या है:Answer /  उत्तर : - 1265SSC Constable (GD) 2013 - 25 June 2013 - Evening shift Explanation in Video / पूर्ण हल जानने के लिए विडिओ देखिए : -   