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Difference between revisions of "2005 AMC 12B Problems/Problem 15"
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== Solution ==
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== Solution 1==
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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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<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>.  
  
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Case 1: (e+f+g+h)=11
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Case 1: <math>(e+f+g+h)=11</math>
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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.
 
  
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Case 2: (e+f+g+h)=21
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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>.
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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
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Case 2: <math>(e+f+g+h)=21</math>
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Checking all the values for <math>e</math>,<math>f</math>,<math>g</math>,and <math>h</math> each individually may be time-consuming, instead of only having <math>1</math> solution like Case 1. We can try a different approach by looking at <math>(a+b+c+d)</math> first. If <math>(e+f+g+h)=21</math>, <math>10(a+b+c+d)=221-21=200</math>, or <math>(a+b+c+d)=20</math>. That means <math>(a+b+c+d)+(e+f+g+h)=21+20=41</math>. We know <math>(1+2+3+4+5+6+7+8+9)=45</math>, so the missing digit is <math>45-41=\boxed{\mathrm{(D)}\ 4}</math>
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== Solution 2 ==
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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 ==
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* [[2005 AMC 12B Problems]]
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{{AMC12 box|year=2005|ab=B|num-b=14|num-a=16}}
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{{MAA Notice}}

Latest revision as of 20:02, 21 December 2020

Problem

The sum of four two-digit numbers is $221$. None of the eight digits is $0$ and no two of them are the same. Which of the following is not included among the eight digits?

$\mathrm{(A)}\ 1 \qquad \mathrm{(B)}\ 2 \qquad \mathrm{(C)}\ 3 \qquad \mathrm{(D)}\ 4 \qquad \mathrm{(E)}\ 5$

Solution 1

$221$ can be written as the sum of four two-digit numbers, let's say $\overline{ae}$, $\overline{bf}$, $\overline{cg}$, and $\overline{dh}$. 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 none of them work. Therefore, $(e+f+g+h)$ cannot add up to $11$.

Case 2: $(e+f+g+h)=21$

Checking all the values for $e$,$f$,$g$,and $h$ 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 $45-41=\boxed{\mathrm{(D)}\ 4}$

Solution 2

Alternatively, we know that a number is congruent to the sum of its digits mod 9, so $221 \equiv 5 \equiv 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 - d \equiv -d$, where $d$ is some digit. Clearly, $\boxed{d = 4}$.

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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