Answer: Let's call the total number of children in the football club "x".
According to the problem:
When the children are put in rows of 9, there are 2 children left over. This means that x is 2 more than a multiple of 9, or in other words: x ≡ 2 (mod 9).
When the children are put in rows of 12, there are 2 children left over. This means that x is also 2 more than a multiple of 12: x ≡ 2 (mod 12).
To find the least value of x that satisfies both of these congruences, we can use the Chinese Remainder Theorem (CRT).
First, let's rewrite the congruences in a standard form:
x ≡ 2 (mod 9) --> x = 9a + 2, for some integer "a"
x ≡ 2 (mod 12) --> x = 12b + 2, for some integer "b"
Now we can use CRT to solve for x:
Find the greatest common divisor (GCD) of 9 and 12: gcd(9, 12) = 3.
Divide both sides of each congruence by the GCD:
x/3 ≡ 2/3 (mod 3), where x/3 = 3a + 2/3
x/3 ≡ 2/3 (mod 4), where x/3 = 4b + 2/3
Multiply the first congruence by 4 and the second congruence by -3:
4(x/3 ≡ 2/3 (mod 3)) --> 4(3a + 2/3) ≡ 8/3 (mod 4) --> 4(3a) + 4(2/3) ≡ 2 (mod 4) --> 3a + 1 ≡ 2 (mod 4)
-3(x/3 ≡ 2/3 (mod 4)) --> -3(4b + 2/3) ≡ -2 (mod 3) --> -3(4b) - 3(2/3) ≡ 1 (mod 3) --> b - 1 ≡ 1 (mod 3)
Solve for the remainders in each congruence:
3a + 1 ≡ 2 (mod 4) --> a ≡ 1 (mod 4)
b - 1 ≡ 1 (mod 3) --> b ≡ 2 (mod 3)
Substitute back to find x:
x = 9a + 2 = 9(4k + 1) + 2 = 36k + 11, where k is an integer (since a ≡ 1 (mod 4))
x = 12b + 2 = 12(3j + 2) + 2 = 36j + 26, where j is an integer (since b ≡ 2 (mod 3))
Check that x satisfies both original congruences:
x ≡ 11 (mod 9), which has a remainder of 2 when divided by 9 (as required)
x ≡ 26 (mod 12), which has a remainder of 2 when divided by 12 (as required)
Since we want the least value of x, we can take k = j
Explanation: