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

To solve this problem, we will use the Chinese Remainder Theorem. Here are the steps:

1. First, we need to find 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.

2. The product of the divisors (2, 3, and 7) is 42.

3. Then, we find the individual products: 42/2 = 21, 42/3 = 14, and 42/7 = 6.

4. Next, we find the multiplicative inverse: 211 mod 2 = 1, 142 mod 3 = 1, and 6*1 mod 7 = 1.

5. Now, we multiply the remainder by the individual product and its multiplicative inverse: (1211) + (2142) + (361) = 21 + 56 + 18 = 95.

6. The least number that satisfies the conditions is 95. However, since the product of the divisors (42) is less than 95, we subtract 42 from 95 to get the smallest such number.

7. Therefore, 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 - 42 = 53.