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The children from a football club are put in rows in the sports hall

When put in rows of 9, there are two children left over
When put in rows of 12 there are two children left over

What is the least number if children in the football club

User JSoet
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2 Answers

4 votes

Answer:

38

Explanation:

it would be 38 because we know that the children from a football club are put into rows in the sports hall.

- when putting rows of 9 children, there are 2 children left.

- When put into rows of 12 children, there are 2 left.

LOOK AT THE LCM HERE (LCM OF 9 and 12 = 36) :
9 = 3x3

12 = 2x2x3

= 4x3

= 12

2x2x3x3 = 36

GIVEN that when putting rows of 9 children, there are 2 children left over. Also when put into rows of 12 children, there are 2 left over.

So add the 2 rows with 36 due to there being 2 rows of children LEFT OVER!

36+2=38

The least number of children in the football club = 38.

User Altiano Gerung
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3 votes

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:

User ItzFlubby
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