How to Find a 2-Digit Number Divisible by 2, 3, and 5 with the Same Remainder of 1

When exploring number theory and modular arithmetic, one can encounter intriguing problems like finding a two-digit number that leaves a remainder of 1 when divided by 2, 3, and 5. This article will guide you through the steps to solve such a problem and present a practical example in detail.

Mathematical Setup and Concepts

The problem can be broken down into the following modular equations:

Modular Equations

Mod 2: n ≡ 1 (mod 2)

Mod 3: n ≡ 1 (mod 3)

Mod 5: n ≡ 1 (mod 5)

These conditions imply that the number n - 1 must be divisible by 2, 3, and 5. We can use the least common multiple (LCM) to find the smallest such number.

Step-by-Step Solution

To find the smallest number, let's first calculate the LCM of 2, 3, and 5:

Step 1: Calculate the LCM

LCM(2, 3, 5) 2 * 3 * 5 30

Now, we express n in terms of this LCM:

Step 2: Express n

n - 1 30k for some integer k

n 30k 1

Since we want n to be a two-digit number (10 to 99), we can set up the following inequality:

Step 3: Find Suitable Values of k

10 ≤ 30k 1 ≤ 99

9 ≤ 30k ≤ 98

0.3 ≤ k ≤ 3.3

Since k must be an integer, the possible values for k are 1, 2, and 3:

Step 4: Calculate n for Integer Values of k

k 1 → n 30 * 1 1 31

k 2 → n 30 * 2 1 61

k 3 → n 30 * 3 1 91

Verification

To verify, we can check the remainders of these numbers when divided by 2, 3, and 5:

31 % 2 1, 31 % 3 1, 31 % 5 1

61 % 2 1, 61 % 3 1, 61 % 5 1

91 % 2 1, 91 % 3 1, 91 % 5 1

Hence, the valid two-digit numbers are 31, 61, and 91.

Conclusion

By understanding and applying modular arithmetic and the concept of LCM, we can determine the two-digit number that meets the given conditions. These steps and examples provide a foundational approach to solving similar number theory problems.