The question asks for the smallest number that needs to be added to 91673 to make the sum perfectly divisible by 88. To solve this, we need to find the remainder when 91673 is divided by 88.
A number is divisible by 88 if it is divisible by both 8 and 11, as 8 and 11 are coprime factors of 88 ($8 \times 11 = 88$). We can find the remainder when 91673 is divided by 88 directly.
Let's perform the division:
Divide 91673 by 88.
We can use long division or a calculator.
$91673 \div 88$
So, we can write the division as:
$91673 = 88 \times 1041 + 65$
The remainder when 91673 is divided by 88 is 65.
We want to find the smallest number, let's call it '$x$', such that $91673 + x$ is exactly divisible by 88.
This means that $(91673 + x)$ should have a remainder of 0 when divided by 88.
We know that $91673$ has a remainder of 65 when divided by 88.
So, $(65 + x)$ must be the next multiple of 88, or 0.
To make the sum divisible by 88, we need to add a number that, when added to the current remainder (65), results in the next multiple of 88.
The next multiple of 88 after 65 is 88 itself.
Therefore, we need $65 + x = 88$.
Solving for $x$:
$x = 88 - 65$
$x = 23$
Let's check if adding 23 to 91673 results in a number divisible by 88.
New sum = $91673 + 23 = 91696$
Now, divide 91696 by 88:
$91696 \div 88 = 1042$
Since the division results in a whole number (1042) with no remainder, the sum 91696 is perfectly divisible by 88.
The smallest number that should be added to 91673 is 23.
What smallest number should be added to 16809 so that the sum is completely divisible by 41?