Find the least number which when divided by 5, 6, 3 and 4 leaves a remainder 2, but when divided by 7 leaves no remainder.2660

The correct option is 182.

To find the least number that leaves a remainder of 2 when divided by 5, 6, 3, and 4, but is completely divisible by 7 (leaves no remainder), we follow these step-by-step logical reasoning and mathematical steps:

Step 1: Find the Least Common Multiple (LCM) of the divisors (5, 6, 3, and 4).
First, let's find the prime factorization of each number:
• 5 = 5¹
• 6 = 2¹ × 3¹
• 3 = 3¹
• 4 = 2²

The LCM is obtained by taking the highest power of each prime factor involved:
$\text{LCM}(5,6,3,4)=2^2\times 3^1\times 5^1=4\times 3\times 5=60$

Step 2: Formulate the general expression for the required number.
Any number that leaves a remainder of 2 when divided by 5, 6, 3, and 4 must be of the form:
$\text{Required Number}=60k+2$
where $k$ is a positive integer ($k=1,2,3,...$).

Step 3: Find the value of $k$ such that $60k+2$ is divisible by 7.
We can rewrite 60 in terms of multiples of 7:
$60=7\times 8+4$

So, the expression becomes:
$60k+2=(56k+4k)+2=56k+(4k+2)$

Since $56k$ is always divisible by 7 (as 56 = 7 × 8), the entire number $60k+2$ will be divisible by 7 if and only if $(4k+2)$ is divisible by 7.

Now, test integer values for $k$ starting from 1:
• For $k=1$: $4\left(1\right)+2=6$ (not divisible by 7)
• For $k=2$: $4\left(2\right)+2=10$ (not divisible by 7)
• For $k=3$: $4\left(3\right)+2=14$ (which is completely divisible by 7, as 14 = 7 × 2)

Step 4: Calculate the required number.
Substitute $k=3$ back into the general expression:
$\text{Required Number}=60\left(3\right)+2=180+2=182$

Thus, the least number satisfying all the given conditions is 182.