Find the Least Number Divisible by Given Numbers with a Specific Remainder

Mathematics often challenges us with intriguing problems that require us to find specific numbers with certain properties. One such problem is to determine the least number that, when divided by 6, 7, 8, 9, and 12, leaves a remainder of 1 in each case. This article will guide you through the process of finding such a number and explain the mathematical concepts involved.

The Mathematical Approach

Let the number be N. The condition that N leaves a remainder of 1 when divided by 6, 7, 8, 9, and 12 can be expressed as follows:

N ≡ 1 (mod 6)
N ≡ 1 (mod 7)
N ≡ 1 (mod 8)
N ≡ 1 (mod 9)
N ≡ 1 (mod 12)

This can be rephrased as:

N - 1 ≡ 0 (mod 6)
N - 1 ≡ 0 (mod 7)
N - 1 ≡ 0 (mod 8)
N - 1 ≡ 0 (mod 9)
N - 1 ≡ 0 (mod 12)

Therefore, (N - 1) must be a common multiple of 6, 7, 8, 9, and 12. We can find the least common multiple (LCM) of these numbers to solve for N.

Prime Factorization and LCM Calculation

To find the LCM, we start by determining the prime factorization of each number:

6 2 × 3
7 7
8 2^3
9 3^2
12 2^2 × 3

The LCM is found by taking the highest power of each prime that appears in the factorizations:

For 2: the highest power is 2^3 from 8. For 3: the highest power is 3^2 from 9. For 7: the highest power is 7^1 from 7.

Thus, the LCM is:

LCM 2^3 × 3^2 × 7 8 × 9 × 7

Calculating this step-by-step:

8 × 9 72 72 × 7 504

Therefore, N - 1 504, which means:

N 504 1 505

The least number that meets the criteria is 505.

Explanation and Verification

To further verify, we check if 505 divides into 6, 7, 8, 9, and 12 with a remainder of 1:

505 ÷ 6 84...1 505 ÷ 7 72...1 505 ÷ 8 63...1 505 ÷ 9 56...1 505 ÷ 12 42...1

Hence, the least number that leaves a remainder of 1 when divided by 6, 7, 8, 9, and 12 is indeed 505.

Conclusion

This process involves finding the LCM of the given numbers and then adding 1 to the result to find the least number that satisfies the given conditions. Understanding the prime factorization and LCM is crucial for solving such problems efficiently.