Find the greatest number of four digits which when divided by 10, 15, 21 and 28 leaves 4, 9, 15 and 22 as remainders respectively?A.    9654B.     9666 C.     9664D.    9864

To solve this problem, we need to find a number that when subtracted by 4, 9, 15, and 22, is divisible by 10, 15, 21, and 28 respectively.

Step 1: Since the number is a four-digit number, the maximum it can be is 9999.

Step 2: We subtract the largest remainder from 9999, which is 22. So, 9999 - 22 = 9977.

Step 3: Now, we need to find a number that is less than or equal to 9977 and is divisible by 10, 15, 21, and 28. The least common multiple (LCM) of 10, 15, 21, and 28 is 420.

Step 4: Divide 9977 by 420, we get a quotient of 23 and a remainder of 377.

Step 5: Subtract the remainder from 9977, we get 9977 - 377 = 9600. This is the largest number less than 9977 that is divisible by 10, 15, 21, and 28.

Step 6: Now, add back the remainders to 9600. When we add 4, we get 9604, which is not a four-digit number divisible by 10, 15, 21, and 28. When we add 9, we get 9609, which is not a four-digit number divisible by 10, 15, 21, and 28. When we add 15, we get 9615, which is not a four-digit number divisible by 10, 15, 21, and 28. When we add 22, we get 9622, which is a four-digit number divisible by 10, 15, 21, and 28.

So, the greatest number of four digits which when divided by 10, 15, 21 and 28 leaves 4, 9, 15 and 22 as remainders respectively is 9622. However, this option is not given in the choices. There seems to be a mistake in the options provided.