A is a set of positive integers such that when divided by 2, 3, 4, 5 and 6 leaves the reminder 1, 2, 3, 4 and 5 respectively. How many integers between 0 and 100 belong to the set A?
One
Finding Positive Integers with Specific Remainders
The problem asks us to find the count of positive integers, let's call the set of these integers A, between 0 and 100 that satisfy specific conditions related to division and remainders. An integer \(x\) belongs to set A if, when divided by 2, 3, 4, 5, and 6, it leaves remainders 1, 2, 3, 4, and 5, respectively.
Understanding the Remainder Conditions
We can write the given conditions using modular arithmetic notation (congruences):
- When divided by 2, the remainder is 1: \(x \equiv 1 \pmod{2}\)
- When divided by 3, the remainder is 2: \(x \equiv 2 \pmod{3}\)
- When divided by 4, the remainder is 3: \(x \equiv 3 \pmod{4}\)
- When divided by 5, the remainder is 4: \(x \equiv 4 \pmod{5}\)
- When divided by 6, the remainder is 5: \(x \equiv 5 \pmod{6}\)
Let's observe the relationship between the divisor and the remainder in each condition:
- \(2 - 1 = 1\)
- \(3 - 2 = 1\)
- \(4 - 3 = 1\)
- \(5 - 4 = 1\)
- \(6 - 5 = 1\)
In every case, the difference between the divisor and the remainder is 1. This pattern is very helpful. It means that if we add 1 to the integer \(x\), the new number, \(x+1\), will be perfectly divisible by each of the divisors: 2, 3, 4, 5, and 6.
Using the Least Common Multiple (LCM)
If a number is divisible by several numbers, it must be divisible by their Least Common Multiple (LCM). In this case, \(x+1\) is divisible by 2, 3, 4, 5, and 6. Therefore, \(x+1\) must be a multiple of LCM(2, 3, 4, 5, 6).
Let's calculate the LCM of 2, 3, 4, 5, and 6:
- Prime factorization of 2 is 2.
- Prime factorization of 3 is 3.
- Prime factorization of 4 is \(2^2\).
- Prime factorization of 5 is 5.
- Prime factorization of 6 is \(2 \times 3\).
To find the LCM, we take the highest power of all prime factors involved: \(2^2\), 3, and 5.
LCM(2, 3, 4, 5, 6) \( = 2^2 \times 3 \times 5 = 4 \times 3 \times 5 = 60\).
So, \(x+1\) must be a multiple of 60. We can write this as:
\(x+1 = 60k\)
where \(k\) is a positive integer (since \(x\) is a positive integer, \(x+1\) must also be positive, and 60 is positive, so \(k\) must be positive). From this, we can express \(x\) as:
\(x = 60k - 1\)
Finding Integers in the Range (0, 100)
We are looking for positive integers \(x\) such that \(0 < x < 100\). We substitute our expression for \(x\) into this inequality:
\(0 < 60k - 1 < 100\)
Now, we solve for the possible integer values of \(k\).
First inequality: \(0 < 60k - 1\)
\(1 < 60k\)
\(k > \frac{1}{60}\)
Since \(k\) must be a positive integer, the smallest possible integer value for \(k\) is 1.
Second inequality: \(60k - 1 < 100\)
\(60k < 101\)
\(k < \frac{101}{60}\)
\(k < 1.6833...\)
So, we need to find integer values of \(k\) such that \(k > \frac{1}{60}\) and \(k < \frac{101}{60}\). The only integer that satisfies both conditions is \(k=1\).
Calculating the Integer x
Using the only possible value of \(k=1\), we find the value of \(x\):
\(x = 60(1) - 1 = 60 - 1 = 59\)
Verifying the Solution
Let's check if \(x=59\) meets all the original conditions and is within the specified range:
- Is 59 between 0 and 100? Yes, \(0 < 59 < 100\).
- Is 59 a positive integer? Yes.
- 59 divided by 2: \(59 = 2 \times 29 + 1\) (Remainder 1) - Correct.
- 59 divided by 3: \(59 = 3 \times 19 + 2\) (Remainder 2) - Correct.
- 59 divided by 4: \(59 = 4 \times 14 + 3\) (Remainder 3) - Correct.
- 59 divided by 5: \(59 = 5 \times 11 + 4\) (Remainder 4) - Correct.
- 59 divided by 6: \(59 = 6 \times 9 + 5\) (Remainder 5) - Correct.
All conditions are satisfied for \(x=59\).
Since only \(k=1\) yielded a solution within the range, there is only one integer between 0 and 100 that belongs to the set A.
Therefore, the number of such integers is one.
Revision Table: Understanding Remainders
| Condition | Mathematical Form | Implies Divisibility of \(x+1\) by |
|---|---|---|
| \(x\) divided by 2 leaves remainder 1 | \(x \equiv 1 \pmod{2}\) | 2 |
| \(x\) divided by 3 leaves remainder 2 | \(x \equiv 2 \pmod{3}\) | 3 |
| \(x\) divided by 4 leaves remainder 3 | \(x \equiv 3 \pmod{4}\) | 4 |
| \(x\) divided by 5 leaves remainder 4 | \(x \equiv 4 \pmod{5}\) | 5 |
| \(x\) divided by 6 leaves remainder 5 | \(x \equiv 5 \pmod{6}\) | 6 |
Additional Information: Chinese Remainder Theorem and Congruences
The problem we solved is a classic example of a system of linear congruences. Such systems can often be solved using the Chinese Remainder Theorem (CRT). The CRT states that if we have a system of congruences \(x \equiv a_i \pmod{n_i}\) where the moduli \(n_i\) are pairwise coprime (their greatest common divisor is 1), then there is a unique solution modulo the product of the moduli (\(n_1 n_2 ... n_k\)).
In our case, the moduli are 2, 3, 4, 5, and 6. These moduli are not pairwise coprime (e.g., GCD(2, 4) = 2, GCD(4, 6) = 2, GCD(2, 6) = 2, GCD(3, 6) = 3). However, we can still solve this system. The structure of the remainders having a constant difference with the moduli (\(n_i - a_i = 1\)) allowed us to simplify the system to a single congruence \(x \equiv -1 \pmod{\text{LCM}(2,3,4,5,6)}\), which is \(x \equiv -1 \pmod{60}\) or \(x \equiv 59 \pmod{60}\).
This means that any integer \(x\) satisfying all the conditions must be of the form \(x = 60k + 59\) for some integer \(k\). This is equivalent to \(x = 60(k+1) - 1\). If we let \(K = k+1\), then \(x = 60K - 1\). Since we are looking for positive integers \(x\), and \(x = 60K - 1\), if \(K\) is a positive integer, \(x\) will also be positive (\(60(1)-1 = 59 > 0\)). So \(K\) corresponds to the \(k\) in our solution derivation, and it must be a positive integer.
The general solution \(x \equiv 59 \pmod{60}\) represents all integers satisfying the conditions. We then filtered this general solution to find integers within the specific range (0, 100).
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select the correct answer using the code given below:
