Find the least number which by decreasing 20 os exactly divisible by 18, 21, 28 and 30. About the author Skylar
Answer: First , we find the least number which is exactly divisible by the numbers 18,21,28 and 30 . For this , we find the L.C.M of 18,21,28 and 30. ∴ L.C.M =2×2×3×3×5×7=1260 According to given , the required will be 20 more than 1260. The required number =1260+20=1280 Reply
Answer: 1280 Step-by-step explanation: First , we find the least number which is exactly divisible by the numbers 18,21,28 and 30 . For this , we find the L.C.M of 18,21,28 and 30. ∴ L.C.M =2×2×3×3×5×7=1260 According to given , the required will be 20 more than 1260. The required number =1260+20=1280 Reply
Answer:
First , we find the least number which is exactly divisible by the numbers 18,21,28 and 30 . For this , we find the L.C.M of 18,21,28 and 30.
∴ L.C.M =2×2×3×3×5×7=1260
According to given , the required will be 20 more than 1260.
The required number =1260+20=1280
Answer: 1280
Step-by-step explanation:
First , we find the least number which is exactly divisible by the numbers 18,21,28 and 30 . For this , we find the L.C.M of 18,21,28 and 30.
∴ L.C.M =2×2×3×3×5×7=1260
According to given , the required will be 20 more than 1260.
The required number =1260+20=1280