+864 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_
+135 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≡2 (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≡1 (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.
+135 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≡2 (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≡1 (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.
