Find the smallest number such that when divided by 12, 18, 21 and 28, it leaves remainder 3 in each case?a.1012b.1010c.1011d.1009

The smallest number that leaves a remainder of 3 when divided by 12, 18, 21, and 28 is found by:

Step 1: Find the least common multiple (LCM) of the four numbers. The LCM of 12, 18, 21, and 28 is 252.

Step 2: Add the remainder to the LCM. The remainder is 3, so 252 + 3 = 255.

Therefore, the smallest number that leaves a remainder of 3 when divided by 12, 18, 21, and 28 is 255. However, this number is not in the options given.

Let's check the options:

a. 1012 divided by 12, 18, 21, and 28 leaves a remainder of 12, 6, 15, and 4 respectively. b. 1010 divided by 12, 18, 21, and 28 leaves a remainder of 10, 4, 13, and 2 respectively. c. 1011 divided by 12, 18, 21, and 28 leaves a remainder of 11, 5, 14, and 3 respectively. d. 1009 divided by 12, 18, 21, and 28 leaves a remainder of 9, 3, 12, and 1 respectively.

None of the options given leaves a remainder of 3 when divided by 12, 18, 21, and 28. There seems to be a mistake in the question or the options provided.