Question 9 - Figure it out - Page 110, 111 - Chapter 5 Class 6 - Prime Time (Ganita Prakash)

Transcript

Question 9 Find the smallest number that is a multiple of all the numbers from 1 to 10, except for 7.Here is a step-by-step way to build that number: Start with the largest number The number must be a multiple of 10. So, let's start with 10. The factors of 10 are 2 and 5. This means our final answer must be divisible by 2 and 5. Consider the next number 9 Our number must also be a multiple of 9. The factors of 9 are 3 & 3. Our current number (10) is not divisible by 9. To make it divisible by 9, we need to include its factors (3 × 3). So far, our number needs the factors: 2 × 5 × 3 × 3 = 90 Consider 8 Our number must be a multiple of 8. The factors of 8 are 2 × 2 × 2. Our current number (90) only has one factor of 2. To make it divisible by 8, it needs to have three factors of 2. We'll replace our single 2 with three 2s. Our factors are now: (2 × 2 × 2) × 3 × 3 × 5 = 360 Check the remaining numbers: Now we see if our new number, 360, is a multiple of the rest of the numbers on the list Is 360 a multiple of 6? 6 is 2 × 3. Our number has plenty of 2s and 3s. Yes, 360 ÷ 6 = 60. Is 360 a multiple of 5? Yes, it ends in 0. 360 ÷ 5 = 7 Is 360 a multiple of 4? 4 is 2 × 2. Our number has three 2s. Yes, 360 ÷ 4 = 90. Is 360 a multiple of 3? Yes, it has two 3s as factors. 360 ÷ 3 = 120. Is 360 a multiple of 2? Yes, it's an even number. 360 ÷ 2 = 180. Is 360 a multiple of 1? Yes, all whole numbers are. Since 360 is the smallest number we could build that satisfies all these conditions, the answer is 360.