A school orders $99$ textbooks, all for the same price. When the bill for the total order comes, the first and last digits are obscured. What are the missing digits?
_18,486.7_
Let's rewrite the number as \(a18,486.7b\). This number must be both divisible by 9 and 11. For the number to be divisible by 9, the digits must add up to a multile of 9. The sum of all the digits are \(34 + a + b\). The closest multiple of 9 is 36. Therefore, \(a+b \equiv 2\text{ } (\text{mod }9)\), because the sum of any 2 one-digit numbers can't be any greater than 18. For a number to be divisible by 11, the result of alternating adding and subtracting the digits must be divisible by 11 (ex: 121, 1 - 2 + 1 = 0, so 121 is divisible by 11). Therefore, \(a + 1 - 8 + 4 - 8 + 6 - 7 + b = -12+a+b\)
The closest multiple of 11 is -11, so \(a+b\equiv1\text{ }(\text{mod 11} )\). We can now narrow down the possibilities. No two single digit numbers can add up to more than 18. a + b can now only be 2 and 11 for the first equation, and a + b can now only be 1 and 12 for the second equation. Since none of these are the same, there are no possibilities for the missing digits.
Let's rewrite the number as \(a18,486.7b\). This number must be both divisible by 9 and 11. For the number to be divisible by 9, the digits must add up to a multile of 9. The sum of all the digits are \(34 + a + b\). The closest multiple of 9 is 36. Therefore, \(a+b \equiv 2\text{ } (\text{mod }9)\), because the sum of any 2 one-digit numbers can't be any greater than 18. For a number to be divisible by 11, the result of alternating adding and subtracting the digits must be divisible by 11 (ex: 121, 1 - 2 + 1 = 0, so 121 is divisible by 11). Therefore, \(a + 1 - 8 + 4 - 8 + 6 - 7 + b = -12+a+b\)
The closest multiple of 11 is -11, so \(a+b\equiv1\text{ }(\text{mod 11} )\). We can now narrow down the possibilities. No two single digit numbers can add up to more than 18. a + b can now only be 2 and 11 for the first equation, and a + b can now only be 1 and 12 for the second equation. Since none of these are the same, there are no possibilities for the missing digits.