Difference between revisions of "2005 AMC 12B Problems/Problem 15"
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== Solution == | == Solution == | ||
| + | 221 can be written as the sum of eight two-digit numbers, lets say ae, bf, cg, and dh. Then 221= 10(a+b+c+d)+(e+f+g+h). The last digit of 221 is 1, and 10(a+b+c+d) won't affect the units digits, so (e+f+g+h) must end with 1. The smallest value (e+f+g+h) can have is (1+2+3+4)=10, and the greatest value is (6+7+8+9)=30. Therefore, (e+f+g+h) must equal 11 or 21. | ||
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| + | Case 1: (e+f+g+h)=11 | ||
| + | The only distinct positive integers that can add up to 11 is (1+2+3+5). So, a,b,c, and d must include four of the five numbers (4,6,7,8,9). We have 10(a+b+c+d)=221-11=210, or a+b+c+d=21. We can add all of 4+6+7+8+9=34, and try subtracting one number to get to 21, but to no avail. Therefore, (e+f+g+h) cannot add up to 11. | ||
| + | |||
| + | Case 2: (e+f+g+h)=21 | ||
| + | Checking all the values for e,f,g,h each individually may be time-consuming, instead of only having 1 solution like case 1. We can try a different approach by looking at (a+b+c+d) first. If (e+f+g+h)=21, 10(a+b+c+d)=221-21=200, or (a+b+c+d)=20. That means (a+b+c+d)+(e+f+g+h)=21+20=41. We know (1+2+3+4+5+6+7+8+9)=45, so the missing digit is 45-41=4 | ||
== See also == | == See also == | ||
* [[2005 AMC 12B Problems]] | * [[2005 AMC 12B Problems]] | ||
Revision as of 19:09, 11 September 2010
Problem
The sum of four two-digit numbers is 
. None of the eight digits is 
and no two of them are the same. Which of the following is not included among the eight digits?
Solution
221 can be written as the sum of eight two-digit numbers, lets say ae, bf, cg, and dh. Then 221= 10(a+b+c+d)+(e+f+g+h). The last digit of 221 is 1, and 10(a+b+c+d) won't affect the units digits, so (e+f+g+h) must end with 1. The smallest value (e+f+g+h) can have is (1+2+3+4)=10, and the greatest value is (6+7+8+9)=30. Therefore, (e+f+g+h) must equal 11 or 21.
Case 1: (e+f+g+h)=11 The only distinct positive integers that can add up to 11 is (1+2+3+5). So, a,b,c, and d must include four of the five numbers (4,6,7,8,9). We have 10(a+b+c+d)=221-11=210, or a+b+c+d=21. We can add all of 4+6+7+8+9=34, and try subtracting one number to get to 21, but to no avail. Therefore, (e+f+g+h) cannot add up to 11.
Case 2: (e+f+g+h)=21 Checking all the values for e,f,g,h each individually may be time-consuming, instead of only having 1 solution like case 1. We can try a different approach by looking at (a+b+c+d) first. If (e+f+g+h)=21, 10(a+b+c+d)=221-21=200, or (a+b+c+d)=20. That means (a+b+c+d)+(e+f+g+h)=21+20=41. We know (1+2+3+4+5+6+7+8+9)=45, so the missing digit is 45-41=4
