How to Find a Number That Divides by a Set of Numbers with a Remainder of 1 and by 11 with No Remainder
Number theory is a fascinating field that explores the properties of numbers and their relationships. One of the more intriguing problems involves finding a number that leaves a specific remainder when divided by some numbers but is divisible by another without any remainder. In this article, we will explore this problem in detail and provide a solution using modular arithmetic and the least common multiple (LCM).
Problem Statement
The problem we are solving is to find a number N that, when divided by 2, 3, 4, 5, 6, 7, 8, 9, and 10, leaves a remainder of 1, and when divided by 11, leaves no remainder (i.e., it is divisible by 11).
Step-by-Step Solution
Step 1: Finding the Number Leaving a Remainder of 1
To find a number that leaves a remainder of 1 when divided by 2, 3, 4, 5, 6, 7, 8, 9, and 10, we first find the least common multiple (LCM) of these numbers. The LCM represents the smallest number that is a multiple of all these values.
Calculating the LCM:
LCM(2, 3, 4, 5, 6, 7, 8, 9, 10) 2520.
So, we need a number that looks like 2520k 1, where k is an integer.
Step 2: Ensuring Divisibility by 11
We need to ensure that the number N 2520k 1 is divisible by 11. In other words, we need 2520k 1 {} 0 mod 11.
Let's compute 2520 mod 11:
2520 div 11 229 remainder 1, so 2520 equiv 1 mod 11.
Hence, the equation simplifies to:
1 cdot k 1 equiv 0 mod 11, which implies k equiv -1 mod 11, or k equiv 10 mod 11.
So, we can write k 11m 10, where m is an integer.
Substituting k back into the equation N 2520k 1:
N 2520(11m 10) 1 27720m 25200 1 27720m 25201.
Therefore, the general solution for N is:
N 27720m 25201, for any integer m.
To find the smallest positive solution, set m 0 to get N 25201.
Conclusion
The smallest number that meets all the conditions is 25201. This number leaves a remainder of 1 when divided by 2, 3, 4, 5, 6, 7, 8, 9, and 10, and is divisible by 11 without any remainder.
Additional Examples
Example 1: Solving x 1 mod 2, x 1 mod 3, ..., x 1 mod 6, x 0 mod 11
In this case, we need to combine the last two equations. Let's start by finding the LCM of 2 through 6:
LCM(2, 3, 4, 5, 6) 60, and combine it with x 0 mod 11 to get:
x 1 mod 6 0 mod 11
Multiplying the first equation by 11, we get:
11x 11 mod 660.
Now, solve for x modulo 600 (since 11 times 60 equals 660, and we can use 600 for simplification):
11x equiv 10 mod 660, or 6x equiv 1 mod 11, or x equiv 10 mod 11.
Hence, x 121 is a solution.
Example 2: Solving x 1 mod 2, x 1 mod 3, ..., x 1 mod 10, x 0 mod 11
In this scenario, we simplify the problem by removing redundant conditions, focusing on x 1 mod 9, x 1 mod 8, x 1 mod 7, x 1 mod 5, and x 0 mod 11.
Since the LCM of 9, 8, 7, and 5 is 2520, we get:
x 1 mod 2520, x 0 mod 11.
Combining these, we get:
x 2520k 1 must be divisible by 11.
Calculating 2520 mod 11,
2520 div 11 229 remainder 1, so 2520 equiv 1 mod 11.
The equation becomes:
1 cdot k 1 equiv 0 mod 11, or k equiv 10 mod 11, hence k 11m 10.
Substituting k back into the equation:
x 2520(11m 10) 1 25200 27720m 1 27720m 25201.
Thus, the solution is:
x 25201.
Therefore, the smallest solution is x 25201 for the combined conditions.