2018 AMC 8 Problem 14

Below is the video solution and professionally curated solution for Problem 14 of the 2018 AMC 8, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2018 AMC 8 solutions, or check the answer key.

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Concepts:digitsoptimization

Difficulty rating: 1140

14.

Let NN be the greatest five-digit number whose digits have a product of 120.120. What is the sum of the digits of N?N?

15 15

16 16

17 17

18 18

20 20

Video solution:
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Written solution:

To make the largest possible 55 digit number, we must maximize the first digit (the digit in the ten-thousands place).

The largest number that is strictly less than 1010 and divides 120120 is 8,8, so the first digit must be 8.8. Therefore, the product of the remaining number is 15.15.

Similarly, we must now maximize the second digit.

The largest number that is less than 1010 and divides 1515 is 5,5, so the second digit is 5.5. Therefore, the product of the remaining number is 3.3.

We must then maximize the third digit.

The largest number that is less than 1010 and divides 33 is 3,3, so the third digit is 3.3. Therefore, the product of the remaining number is 1.1. This means the 4th and 5th digits are 1.1.

This makes N=85311,N = 85311, so the sum of the digits is 8+5+3+1+1=188+5+3+1+1=18

Thus, D is the correct answer.

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