Saved Bookmarks
| 1. |
Find the smallest number which when increased by 17 is exactly divisible by both 520and 468 |
| Answer» The smallest number which when increased by 17 is exactly divisible by both 520 and 468 is obtained by subtracting 17 from the LCM of 520 and 468.LCM( 520 , 468) = 2 × 2 × 2 × 3 × 3 × 5 × 13 = 4680.Smallest number which when increased by 17 is exactly divisible by both 520 and 468 So the no. is 4680 – 17 = 4663 | |