Finding the Least Number with a Specific Remainder

This problem asks for the smallest positive integer that leaves a remainder of 1 when divided by 7, 12, and 15. This is a common type of problem that involves finding the Least Common Multiple (LCM).

Understanding the Remainder Concept

If a number leaves a remainder of 1 when divided by 7, 12, and 15, it means that if we subtract 1 from that number, the result will be perfectly divisible by 7, 12, and 15.

Let the required number be $N$. According to the problem:

• $N = 7 \times k + 1$ for some integer $k$
• $N = 12 \times m + 1$ for some integer $m$
• $N = 15 \times n + 1$ for some integer $n$

This implies that $N - 1$ is a multiple of 7, 12, and 15.

Therefore, $N - 1$ must be the Least Common Multiple (LCM) of 7, 12, and 15.

Calculating the Least Common Multiple (LCM)

To find the LCM of 7, 12, and 15, we first find the prime factorization of each number:

• Prime factorization of 7: $7$
• Prime factorization of 12: $2 \times 2 \times 3 = 2^2 \times 3$
• Prime factorization of 15: $3 \times 5$

The LCM is found by taking the highest power of all prime factors that appear in any of the factorizations:

$ \text{LCM}(7, 12, 15) = 2^2 \times 3^1 \times 5^1 \times 7^1 $

Calculating the value:

$ \text{LCM}(7, 12, 15) = 4 \times 3 \times 5 \times 7 $

$ \text{LCM}(7, 12, 15) = 12 \times 35 $

$ \text{LCM}(7, 12, 15) = 420 $

Determining the Required Number

We established that the required number $N$ is one more than the LCM of 7, 12, and 15.

$ N - 1 = \text{LCM}(7, 12, 15) $

$ N - 1 = 420 $

Adding 1 to both sides to find $N$:

$ N = 420 + 1 $

$ N = 421 $

Conclusion

The least number which, when divided by 7, 12, and 15, leaves a remainder of 1 in each case is 421.