To solve this problem, we can use the Chinese Remainder Theorem.
According to the Chinese Remainder Theorem, if we have a system of congruences of the form:
x ≡ a1 (mod m1)
x ≡ a2 (mod m2)
...
x ≡ an (mod mn)
where the mi are pairwise coprime (that is, they have no common factors other than 1), then there exists a unique solution for x (modulo the product of the mi) that satisfies all of the congruences.
In this case, we have the following system of congruences:
x ≡ 3 (mod 5)
x ≡ 4 (mod 6)
x ≡ 1 (mod 7)
Since 5, 6, and 7 are pairwise coprime, we can use the Chinese Remainder Theorem to find a solution for x.
The least possible value of x that satisfies all of the congruences is the one that is congruent to 0 (modulo the product of the mi). In this case, the product of the mi is 5 * 6 * 7 = 210, so we want to find the least value of x that is congruent to 0 (modulo 210).
We can solve for x using the following steps:
Find the inverse of 5 (modulo 6). The inverse of 5 (modulo 6) is 5 because 5 * 5 ≡ 1 (mod 6).
Multiply the inverse from step 1 by 3 (modulo 6) to get x ≡ 15 (modulo 30).
Find the inverse of 6 (modulo 7). The inverse of 6 (modulo 7) is 5 because 6 * 5 ≡ 1 (mod 7).
Multiply the inverse from step 3 by 4 (modulo 7) to get x ≡ 20 (modulo 42).
Find the inverse of 7 (modulo 5). The inverse of 7 (modulo 5) is 3 because 7 * 3 ≡ 1 (mod 5).
Multiply the inverse from step 5 by 1 (modulo 5) to get x ≡ 3 (modulo 35).
Find a solution for x that is congruent to 0 (modulo 210) using the values from steps 2, 4, and 6. We can do this by adding 210 to each of the values until we get a value that is congruent to 0 (modulo 210).
For example, starting with the value from step 2 (15), we can add 210 to get x ≡ 225 (modulo 210).
Starting with the value from step 4 (20), we can add 210 to get x ≡ 230 (modulo 210).
Starting with the value from step 6 (3), we can add 210 to get x ≡ 213 (modulo 210).
Since all three of these values are congruent to 0 (modulo 210), they are all valid solutions for x. The least of these values is 225, so the least possible value of x that satisfies all of the congruences is 225.
Therefore, the answer to the problem is 225.
Answer:
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Let the number be N.
If N is divided by 5 the remainder is 3:
Similarly:
We can observe N + 2 is divisible by both 5 and 6, then it can be represented as:
so
The left side is divisible by 3, then q = 3t:
The least m and t would be m = 5 and t = 7.
Then:
Proof:
