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A woman with a basket of eggs finds that if she removes the eggs from the basket 3 or 5 at a time, there is always 1 egg left. However, if she removes the eggs 7 at a time, there are no eggs left. If the basket holds up to 100 eggs, how many eggs does she have?

She has [tex]$\square$[/tex] eggs.

Sagot :

To solve the problem, we need to find a number of eggs that satisfies the following conditions:
1. When the eggs are removed in groups of 3, there is 1 egg left.
2. When the eggs are removed in groups of 5, there is 1 egg left.
3. When the eggs are removed in groups of 7, there are no eggs left.
4. The total number of eggs is less than or equal to 100.

These conditions can be translated into a set of congruences:
[tex]\[ \text{(i)} \quad N \equiv 1 \pmod{3} \][/tex]
[tex]\[ \text{(ii)} \quad N \equiv 1 \pmod{5} \][/tex]
[tex]\[ \text{(iii)} \quad N \equiv 0 \pmod{7} \][/tex]

### Step-by-step Solution:

1. Let's consider the third condition first; this tells us that the number of eggs [tex]\( N \)[/tex] must be a multiple of 7. Therefore, we can write:
[tex]\[ N = 7k \text{ for some integer } k \][/tex]

2. Next, using the first condition, [tex]\( N \equiv 1 \pmod{3} \)[/tex], we substitute [tex]\( N \)[/tex] and get:
[tex]\[ 7k \equiv 1 \pmod{3} \][/tex]

This simplifies to:
[tex]\[ 7k \mod 3 = 1 \][/tex]
[tex]\[ k \equiv 1 \pmod{3} \][/tex]
[tex]\[ \Rightarrow k = 3m + 1 \text{ for some integer } m \][/tex]

Therefore, substituting back for [tex]\( N \)[/tex]:
[tex]\[ N = 7(3m + 1) \][/tex]
[tex]\[ N = 21m + 7 \][/tex]

3. Now we use the second condition, [tex]\( N \equiv 1 \pmod{5} \)[/tex], to find [tex]\( m \)[/tex]. Substitute [tex]\( N \)[/tex]:
[tex]\[ 21m + 7 \equiv 1 \pmod{5} \][/tex]

Simplifying, we have:
[tex]\[ 21m + 7 \equiv 1 \pmod{5} \][/tex]
[tex]\[ 21m \equiv -6 \pmod{5} \][/tex]
[tex]\[ 21m \equiv -1 \pmod{5} \][/tex]
[tex]\[ 21 \equiv 1 \pmod{5} \][/tex]

Hence:
[tex]\[ m \equiv -1 \pmod{5} \][/tex]
[tex]\[ m = 5n - 1 \text{ for some integer } n \][/tex]

4. Finally, substituting [tex]\( m \)[/tex] back into the equation for [tex]\( N \)[/tex]:
[tex]\[ N = 21(5n - 1) + 7 \][/tex]
[tex]\[ N = 105n - 21 + 7 \][/tex]
[tex]\[ N = 105n - 14 \][/tex]

Since the number of eggs must be less than or equal to 100:
[tex]\[ 105n - 14 \leq 100 \][/tex]
[tex]\[ 105n \leq 114 \][/tex]
[tex]\[ n \leq \frac{114}{105} \][/tex]
[tex]\[ n \leq 1.0857 \][/tex]

Since [tex]\( n \)[/tex] must be an integer, [tex]\( n = 1 \)[/tex]:

Hence,
[tex]\[ N = 105(1) - 14 \][/tex]
[tex]\[ N = 91 \][/tex]

Therefore, the woman has [tex]\(\boxed{91}\)[/tex] eggs.