Subject:
MathAuthor:
jessegonzalesCreated:
1 year agoHere, 21 = 3 × 7 and 33 = 3 × 11
∴ LCM of 21 and 33 = 3 × 7 × 11 = 231
∵ 231 < 1000, we multiply 231 by such a small number that the resulting number will be the smallest number greater than 1000 and divisible by 231.
Again, we find the greatest number less than 12000 and divisible by 231.
So, we can form an Arithmetic Progression whose first term is 1155 and common difference is 231.
Let, 11781 be the nth term of the A.P.
Then, 1155 + (n - 1) × 231 = 11781
⇒ (n - 1) × 231 = 11781 - 1155 = 10626
⇒ n - 1 = 10626 ÷ 231
⇒ n - 1 = 46
⇒ n = 46 + 1
⇒ n = 47
Therefore, there are 47 numbers between 1000 and 12000 which are divisible by both 21 and 33.
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