Question

Mrs. Abdul has a number of tarts. If she gives 3 or 4 tarts to each child, there will be 1 tart left over each time. If she gives 5 tarts to each child, she will have no tarts left over. What is the least number of tarts that Mrs. Abdul could have?

Answer 1

Let the number of tarts that Mrs. Abdul has be denoted by \( N \).

According to the problem:

1. If she gives 3 tarts to each child, there will be 1 tart left over:
\[
N \equiv 1 \mod 3
\]
This means when \( N \) is divided by 3, the remainder is 1.

2. If she gives 4 tarts to each child, there will also be 1 tart left over:
\[
N \equiv 1 \mod 4
\]
This means when \( N \) is divided by 4, the remainder is also 1.

3. If she gives 5 tarts to each child, she will have no tarts left over:
\[
N \equiv 0 \mod 5
\]
This means that \( N \) is completely divisible by 5.

From the first two conditions, we can conclude that:
\[
N - 1 \equiv 0 \mod 3 \quad \text{and} \quad N - 1 \equiv 0 \mod 4
\]
This means \( N - 1 \) is divisible by both 3 and 4. The least common multiple of 3 and 4 is 12. Therefore, we can express \( N - 1 \) as:
\[
N - 1 = 12k \quad \text{for some integer } k
\]
Thus,
\[
N = 12k + 1
\]

Next, substituting this into the third condition:
\[
12k + 1 \equiv 0 \mod 5
\]
This simplifies to:
\[
12k \equiv -1 \mod 5
\]
Since \( 12 \mod 5 \equiv 2 \), we rewrite the equation as:
\[
2k \equiv -1 \mod 5
\]
To solve for \( k \), we can express \(-1\) in terms of positive integers. Since \(-1 \equiv 4 \mod 5\), we need to solve:
\[
2k \equiv 4 \mod 5
\]
Dividing both sides by 2, which is valid because 2 has a multiplicative inverse modulo 5 (specifically, 3, since \( 2 \times 3 \equiv 1 \mod 5 \)), we get:
\[
k \equiv 2 \mod 5
\]
This means \( k \) can be expressed as:
\[
k = 5m + 2 \quad \text{for some integer } m
\]
Substituting back into the equation for \( N \):
\[
N = 12(5m + 2) + 1 = 60m + 24 + 1 = 60m + 25
\]
So, we have:
\[
N = 60m + 25
\]
To find the least value of \( N \), we set \( m = 0 \):
\[
N = 25
\]

To verify:
- For \( N = 25 \):
- Dividing by 3: \( 25 \div 3 = 8 \) remainder \( 1 \) (satisfies \( N \equiv 1 \mod 3 \))
- Dividing by 4: \( 25 \div 4 = 6 \) remainder \( 1 \) (satisfies \( N \equiv 1 \mod 4 \))
- Dividing by 5: \( 25 \div 5 = 5 \) remainder \( 0 \) (satisfies \( N \equiv 0 \mod 5 \))

Thus, all conditions are satisfied. Therefore, the least number of tarts that Mrs. Abdul could have is:
\[
\boxed{25}
\]