Difference between revisions of "2005 AMC 12B Problems/Problem 15"
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− | == Solution | + | == Solution 1== |
− | <math>221</math> can be written as the sum of | + | <math>221</math> can be written as the sum of four two-digit numbers, let's say <math>\overline{ae}</math>, <math>\overline{bf}</math>, <math>\overline{cg}</math>, and <math>\overline{dh}</math>. Then <math>221= 10(a+b+c+d)+(e+f+g+h)</math>. The last digit of <math>221</math> is <math>1</math>, and <math>10(a+b+c+d)</math> won't affect the units digits, so <math>(e+f+g+h)</math> must end with <math>1</math>. The smallest value <math>(e+f+g+h)</math> can have is <math>(1+2+3+4)=10</math>, and the greatest value is <math>(6+7+8+9)=30</math>. Therefore, <math>(e+f+g+h)</math> must equal <math>11</math> or <math>21</math>. |
Case 1: <math>(e+f+g+h)=11</math> | Case 1: <math>(e+f+g+h)=11</math> | ||
− | The only distinct positive integers that can add up to <math>11</math> is <math>(1+2+3+5)</math>. So, <math>a</math>,<math>b</math>,<math>c</math>, and <math>d</math> must include four of the five numbers <math>(4,6,7,8,9)</math>. We have <math>10(a+b+c+d)=221-11=210</math>, or <math>a+b+c+d=21</math>. We can add all of <math>4+6+7+8+9=34</math>, and try subtracting one number to get to <math>21</math>, but | + | The only distinct positive integers that can add up to <math>11</math> is <math>(1+2+3+5)</math>. So, <math>a</math>,<math>b</math>,<math>c</math>, and <math>d</math> must include four of the five numbers <math>(4,6,7,8,9)</math>. We have <math>10(a+b+c+d)=221-11=210</math>, or <math>a+b+c+d=21</math>. We can add all of <math>4+6+7+8+9=34</math>, and try subtracting one number to get to <math>21</math>, but none of them work. Therefore, <math>(e+f+g+h)</math> cannot add up to <math>11</math>. |
Case 2: <math>(e+f+g+h)=21</math> | Case 2: <math>(e+f+g+h)=21</math> | ||
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== Solution 2 == | == Solution 2 == | ||
− | Alternatively, we know that a number is congruent to the sum of its digits | + | Alternatively, we know that a number is congruent to the sum of its digits mod 9, so <math>221 \equiv 5 \equiv 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 - d \equiv -d</math>, where <math>d</math> is some digit. Clearly, <math>\boxed{d = 4}</math>. |
== See also == | == See also == | ||
{{AMC12 box|year=2005|ab=B|num-b=14|num-a=16}} | {{AMC12 box|year=2005|ab=B|num-b=14|num-a=16}} | ||
+ | {{MAA Notice}} |
Latest revision as of 20:02, 21 December 2020
Contents
[hide]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 1
can be written as the sum of four two-digit numbers, let's say , , , and . Then . The last digit of is , and won't affect the units digits, so must end with . The smallest value can have is , and the greatest value is . Therefore, must equal or .
Case 1:
The only distinct positive integers that can add up to is . So, ,,, and must include four of the five numbers . We have , or . We can add all of , and try subtracting one number to get to , but none of them work. Therefore, cannot add up to .
Case 2:
Checking all the values for ,,,and each individually may be time-consuming, instead of only having solution like Case 1. We can try a different approach by looking at first. If , , or . That means . We know , so the missing digit is
Solution 2
Alternatively, we know that a number is congruent to the sum of its digits mod 9, so , where is some digit. Clearly, .
See also
2005 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 14 |
Followed by Problem 16 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
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