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A 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? Explain with reasoning ? ?
Most Upvoted Answer
A trader with a basket of eggs finds that if he sells 3 eggs at a time...
We know that the LCM of 3 and 4 is 12. We have also been given that the maximum capacity of the egg basket is 100.

Lets list down the multiples of 12 till the number reaches 100.
12, 24, 36, 48, 60, 72, 84, 96

To leave a remained with 3 as well as 4, these numbers list above need to be one more; so we get the new numbers as follows:
13, 25, 37, 49, 61, 73, 85, 97.

The given statement also states that when 7 eggs are selected, no eggs remain. That mean we are looking for a number that is divisible by 7.

From 13, 25, 37, 49, 61, 73, 85, 97, we see that only 49 matches all the three criteria.

Hence the total number of eggs in the basket is 49
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A trader with a basket of eggs finds that if he sells 3 eggs at a time...
Problem

A 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?

Solution

Let'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:

  • x = 3a + 1

  • x = 4b + 1



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.
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A 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? Explain with reasoning ? ?
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