A number, when divided by $$ 3,4,5 $$, and 6 , has a remainder of 1 . A 21 B 41 C 61 D 81

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To determine which number among the options $$ 21, 41, 61, $$ and $$ 81 $$ leaves a remainder of 1 when divided by $$ 3, 4, 5, $$ and $$ 6 $$, we can use the concept of modular arithmetic and the Least Common Multiple (LCM).

1. **Find the LCM of the divisors:**
   - The divisors are $$ 3, 4, 5, $$ and $$ 6 $$.
   - LCM of $$ 3 $$ and $$ 6 $$ is $$ 6 $$.
   - LCM of $$ 6 $$ and $$ 4 $$ is $$ 12 $$.
   - LCM of $$ 12 $$ and $$ 5 $$ is $$ 60 $$.

So, the LCM of $$ 3, 4, 5, $$ and $$ 6 $$ is $$ 60 $$.

2. **Formulate the condition:**
   - We need to find numbers $$ x $$ such that:
     $$
     x \equiv 1 \pmod{3}, \quad x \equiv 1 \pmod{4}, \quad x \equiv 1 \pmod{5}, \quad x \equiv 1 \pmod{6}
     $$
   - This means $$ x \equiv 1 \pmod{60} $$, since $$ 60 $$ is the LCM.

3. **Check the options:**
   - **A. 21**
     $$
     21 \div 60 = 0 \text{ remainder } 21 \quad (\text{Not } \equiv 1)
     $$
   - **B. 41**
     $$
     41 \div 60 = 0 \text{ remainder } 41 \quad (\text{Not } \equiv 1)
     $$
   - **C. 61**
     $$
     61 = 60 \times 1 + 1 \quad (\equiv 1 \pmod{60})
     $$
   - **D. 81**
     $$
     81 = 60 \times 1 + 21 \quad (\text{Not } \equiv 1)
     $$

Only **61** satisfies the condition of leaving a remainder of 1 when divided by $$ 3, 4, 5, $$ and $$ 6 $$.

**Answer:** C) 61
The number is 61.

A number, when divided by , and 6 , has a remainder of 1 .
A 21
B 41
C 61
D 81

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A number, when divided by , and 6 , has a remainder of 1 . A 21 B 41 C 61 D 81