Find the least number when successively divided by 2,3, and 7 it leaves reminder 1,2, and 3 respectively.Options6367656175

The problem is asking for the least number that leaves a remainder of 1 when divided by 2, a remainder of 2 when divided by 3, and a remainder of 3 when divided by 7.

To solve this, we can use the Chinese Remainder Theorem.

Step 1: Multiply the divisors together. 237 = 42.

Step 2: Divide the result from step 1 by each of the divisors, and note the results. 42/2 = 21, 42/3 = 14, 42/7 = 6.

Step 3: Find a number that when multiplied by the result from step 2 and then divided by the original divisor, leaves a remainder of 1. This is the multiplicative inverse. For 2, this is 1 (211/2 = 10.5, remainder 1). For 3, this is 2 (142/3 = 9.33, remainder 1). For 7, this is 1 (6*1/7 = 0.857, remainder 1).

Step 4: Multiply the results from step 2 by the remainders given in the problem, and then by the multiplicative inverses found in step 3. 2111 = 21, 1422 = 56, 631 = 18.

Step 5: Add the results from step 4 together. 21 + 56 + 18 = 95.

Step 6: If the result from step 5 is larger than the result from step 1, subtract the result from step 1 from the result from step 5 until you get a number smaller than the result from step 1. In this case, 95 is smaller than 42, so we don't need to do this step.

So, the least number that leaves a remainder of 1 when divided by 2, a remainder of 2 when divided by 3, and a remainder of 3 when divided by 7 is 95.

However, none of the options provided (636, 765, 61, 75) match this result. There may be a mistake in the problem or the options provided.