From a basket of tennis balls when counted in twos there was one extra, counted in threes there were two extra, counted in fours there were three extra,
counted in fives there were four extra, counted in sixes there were five extra, but counted in sevens there were no extras. At least how many tennis balls
were there in the basket?

Question

UpStudy AI · Accepted Answer

To find the smallest number of tennis balls $$ n $$ that satisfies all the given conditions, we can approach the problem using modular arithmetic.

### Given Conditions:
1. $$ n \equiv 1 \pmod{2} $$
2. $$ n \equiv 2 \pmod{3} $$
3. $$ n \equiv 3 \pmod{4} $$
4. $$ n \equiv 4 \pmod{5} $$
5. $$ n \equiv 5 \pmod{6} $$
6. $$ n \equiv 0 \pmod{7} $$

### Observing the Pattern:
Notice that in each of the first five conditions, the remainder is one less than the divisor:
- $$ n \equiv -1 \pmod{2} $$
- $$ n \equiv -1 \pmod{3} $$
- $$ n \equiv -1 \pmod{4} $$
- $$ n \equiv -1 \pmod{5} $$
- $$ n \equiv -1 \pmod{6} $$

This suggests that $$ n + 1 $$ is divisible by each of the numbers 2, 3, 4, 5, and 6.

### Step-by-Step Solution:
1. **Find the Least Common Multiple (LCM)**:
   - The LCM of 2, 3, 4, 5, and 6 is 60.
   - Therefore, $$ n + 1 = 60k $$ for some integer $$ k $$.
   - So, $$ n = 60k - 1 $$.

2. **Incorporate the Last Condition**:
   - $$ n $$ must also be divisible by 7: $$ 60k - 1 \equiv 0 \pmod{7} $$.
   - Simplifying: $$ 60k \equiv 1 \pmod{7} $$.
   - Since $$ 60 \equiv 4 \pmod{7} $$, the equation becomes $$ 4k \equiv 1 \pmod{7} $$.
   - Multiplying both sides by 2 (the modular inverse of 4 modulo 7): $$ 8k \equiv 2 \pmod{7} $$, which simplifies to $$ k \equiv 2 \pmod{7} $$.
   - Therefore, $$ k = 7m + 2 $$ for some integer $$ m $$.

3. **Find the Smallest $$ n $$**:
   - Plugging $$ k = 7m + 2 $$ back into $$ n = 60k - 1 $$:
   $$
   n = 60(7m + 2) - 1 = 420m + 120 - 1 = 420m + 119
   $$
   - The smallest positive $$ n $$ occurs when $$ m = 0 $$:
   $$
   n = 119
   $$

4. **Verification**:
   - $$ 119 \div 2 = 59 $$ remainder **1**.
   - $$ 119 \div 3 = 39 $$ remainder **2**.
   - $$ 119 \div 4 = 29 $$ remainder **3**.
   - $$ 119 \div 5 = 23 $$ remainder **4**.
   - $$ 119 \div 6 = 19 $$ remainder **5**.
   - $$ 119 \div 7 = 17 $$ remainder **0**.

All conditions are satisfied.

### Conclusion:
The smallest number of tennis balls in the basket is **119**.

**Answer:** 119
The smallest number of tennis balls in the basket is 119.

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