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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_

 Jul 7, 2024

Best Answer 

 #1
avatar+88 
+1

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.

 Jul 7, 2024
 #1
avatar+88 
+1
Best Answer

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.

Maxematics Jul 7, 2024

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