The problem asks for the smallest number that, when added to 1000, results in a sum that is perfectly divisible by 15.
First, find the remainder when 1000 is divided by 15.
Using division: $1000 \div 15$ $1000 = 15 \times 66 + 10$ The quotient is 66 and the remainder is 10.
To make 1000 divisible by 15, we need to reach the next multiple of 15. The current number (1000) is 10 more than a multiple of 15 ($15 \times 66$).
The amount needed to reach the next multiple of 15 is calculated by subtracting the remainder from the divisor (15).
Least number to add $= 15 - \text{Remainder}$ $= 15 - 10$ $= 5$
Therefore, the least number which when added to 1000 gives a number exactly divisible by 15 is 5. Adding 5 to 1000 gives 1005, and $1005 \div 15 = 67$, confirming it is exactly divisible.
The remainder in the expression $27\frac{3}{4}$ is:
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