Find the least positive integer that leaves remainders of 1, 2 and 3 when divided by 3, 5 and 7 respectively.
With relatively small numbers like in your problem, we could just "grind out" the answer in the following way:
numbers which leave a remainder of 1
when divided by 3:
4 7 10 13 16 19 22 25 28 31 34 37 40 43 46 49 52 55 ....
numbers which leave a remainder of 2 when divided by 5 :
7 12 17 22 27 32 37 42 47 52 57 ..
numbers which leave a remainder of 3 when divided by 7:
10 17 24 31 38 45 52 59 ...
notice that 52 is the first first, thus the smallest number that satisfies your conditions.
If the numbers had been larger, the above method would not be practical and we would use other methods dealing with modular arithmetic. If interested look up the "Chinese Remainder Algorithm", makes for some real math-fun.
To find the least positive integer that leaves remainders of 1, 2, and 3 when divided by 3, 5, and 7 respectively, we can use the Chinese Remainder Theorem.
The Chinese Remainder Theorem states that if we have a system of congruences of the form:
x ≡ a (mod m)
x ≡ b (mod n)
x ≡ c (mod p)
where m, n, and p are pairwise coprime (no two of them share a common factor), then there exists a unique solution x (mod m * n * p).
In this case, we have:
x ≡ 1 (mod 3)
x ≡ 2 (mod 5)
x ≡ 3 (mod 7)
Since 3, 5, and 7 are pairwise coprime, we can find the solution using the Chinese Remainder Theorem.
First, let's find the value of M, which is the product of the three moduli: M = 3 * 5 * 7 = 105.
Next, for each modulus mi, we find the value of Mi, which is M divided by mi: M1 = 105 / 3 = 35, M2 = 105 / 5 = 21, M3 = 105 / 7 = 15.
Then, we need to find the modular inverse of each Mi modulo mi. The modular inverse of Mi modulo mi is the number x such that (Mi * x) ≡ 1 (mod mi). Using the Extended Euclidean Algorithm, we find: M1 inverse ≡ 2 (mod 3), M2 inverse ≡ 1 (mod 5), M3 inverse ≡ 1 (mod 7).
Finally, we can calculate the value of x by taking the sum of the products of each ai (the remainders) with its corresponding Mi and modular inverse: x = (1 * 35 * 2 + 2 * 21 * 1 + 3 * 15 * 1) mod 105 = 23.
Therefore, the least positive integer that leaves remainders of 1, 2, and 3 when divided by 3, 5, and 7 respectively is 23.