The problem asks for the smallest positive integer that, when added to 1549, results in a sum perfectly divisible by 2, 3, 5, and 7.

Finding the Divisor's LCM

For a number to be divisible by 2, 3, 5, and 7, it must be divisible by their Least Common Multiple (LCM). Since 2, 3, 5, and 7 are prime numbers, their LCM is simply their product.

LCM = $2 \times 3 \times 5 \times 7 = 210$

So, the sum must be a multiple of 210.

Calculating the Remainder

We need to find the remainder when 1549 is divided by the LCM (210).

Divide 1549 by 210:

$1549 \div 210$

Performing the division:

$1549 = (7 \times 210) + 79$

The quotient is 7, and the remainder is 79.

Determining the Number to Add

The current number, 1549, leaves a remainder of 79 when divided by 210. To make the sum exactly divisible by 210, we need to reach the next multiple of 210. This means we need to add the difference between the divisor (210) and the remainder (79).

Number to add = LCM - Remainder

Number to add = $210 - 79 = 131$

Verification

Adding 131 to 1549 gives:

$1549 + 131 = 1680$

Now, check if 1680 is divisible by 210:

$1680 \div 210 = 8$

Since the result is an integer (8), 1680 is exactly divisible by 210. Therefore, 131 is the least number that should be added.