A trader with a basket of eggs finds that if he sells 3 eggs at a time...
ProblemA trader with a basket of eggs finds that if he sells 3 eggs at a time, there is only 1 egg left. If he sells 4 eggs at a time, again 1 egg is left. However if the trader sells 7 eggs at a time, there are no eggs left. If the capacity of the basket is 100 eggs, find how eggs are in the basket?
SolutionLet's assume that there are x eggs in the basket.
From the given information, we can write the following equations:
- x ≡ 1 (mod 3)
- x ≡ 1 (mod 4)
- x ≡ 0 (mod 7)
- x ≤ 100
We can simplify the first two equations to:
where a and b are integers.
Substituting x from the first equation into the second equation, we get:
3a + 1 = 4b + 1
3a = 4b
This means that a is a multiple of 4 and b is a multiple of 3.
Let's assume that a = 4c. Then b = 3d, where c and d are integers.
Substituting these values into the equations for x, we get:
x = 3(4c) + 1 = 12c + 1
x = 4(3d) + 1 = 12d + 1
Since x is also a multiple of 7, we can write:
x = 7e
Now, we can substitute 7e for x in the first two equations and simplify:
- 7e ≡ 1 (mod 3)
- 7e ≡ 1 (mod 4)
Solving these equations gives us e = 1, so x = 7.
However, this solution does not satisfy the last condition that x ≤ 100.
We can find the next solution by adding 84 (the LCM of 3, 4, and 7) to 7, and we get x = 91.
We can keep adding 84 until we get a solution that satisfies x ≤ 100.
The solutions are 7, 91, and 175.
Therefore, the basket can have 7, 91, or 175 eggs in it.