Solution:-

Find the LCM for these three numbers (30, 40 and 60).

30 = 2×3×5.
40 = 2³×5.
60 = 2²×3×5.

LCM = 2³×3×5 ---> 120.

Then, 120+7 = 127. This number leaves remainder 7 when divided by these three numbers.

Hence, the smallest number is 127.

Solution:

To find the smallest number which when divided by 30,40 and 60 leaves the remainder 7 in each case, we need to use the concept of LCM (Least Common Multiple) and Congruence.

- LCM of 30, 40 and 60 is 120. This means that any number that satisfies the given conditions must be a multiple of 120.
- Let's assume that the required number is N. Then we can write:
N ≡ 7 (mod 30)
N ≡ 7 (mod 40)
N ≡ 7 (mod 60)
- We can use the Chinese Remainder Theorem to solve these congruences simultaneously. This gives us:
N ≡ 67 (mod 120)
- Therefore, the smallest number which satisfies the given conditions is 67.

Explanation:

- The given problem involves finding a number that satisfies certain divisibility conditions. Such problems can be solved using the concepts of LCM and Congruence.
- LCM of two or more numbers is the smallest number that is divisible by all of them. In this case, the LCM of 30, 40 and 60 is 120. This means that any number that satisfies the given conditions must be a multiple of 120.
- Congruence is a relation between two numbers that have the same remainder when divided by a certain number. In this case, we are given that the required number leaves a remainder of 7 when divided by 30, 40 and 60. We can express this using the congruence notation as shown above.
- The Chinese Remainder Theorem is a theorem in number theory that gives a solution to a system of simultaneous linear congruences. In this case, we have three congruences that we need to solve simultaneously. The theorem gives us a unique solution modulo the LCM of the moduli (in this case, 120).
- Solving the congruences using the Chinese Remainder Theorem as shown above gives us the required solution, which is 67.