Difference between revisions of "1991 AJHSME Problems/Problem 20"

Revision as of 23:02, 17 April 2017

Problem

In the addition problem, each digit has been replaced by a letter. If different letters represent different digits then $C=$

$[asy] unitsize(18); draw((-1,0)--(3,0)); draw((-3/4,1/2)--(-1/4,1/2)); draw((-1/2,1/4)--(-1/2,3/4)); label("$A$",(0.5,2.1),N); label("$B$",(1.5,2.1),N); label("$C$",(2.5,2.1),N); label("$A$",(1.5,1.1),N); label("$B$",(2.5,1.1),N); label("$A$",(2.5,0.1),N); label("$3$",(0.5,-.1),S); label("$0$",(1.5,-.1),S); label("$0$",(2.5,-.1),S); [/asy]$

$\text{(A)}\ 1 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 5 \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 9$

Solution

From this we have $111A+11B+C=300.$ Clearly, $A<3$ . Since $B,C\leq 9$ , $111A > 201 \Rightarrow A\geq 2.$ Thus, $A=2$ and $11B+C=78$ . From here it becomes clear that $B=7$ and $C=1\rightarrow \boxed{\text{A}}$ .

Solution

Using logic, $a+b+c= 10$ , therefore $b+a+1$ (from the carry over) $= 10$ . so b+a=9 A+1=3

Thus, $A=2$ and $11B+C=78$ . From here it becomes clear that $B=7$ and $C=1\rightarrow \boxed{\text{A}}$ .

1991 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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 AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 24: / Line 24: @@
 ==Solution==
-Using logic, a+b+c= 10, therefore b+a+1(from the carry over)= 10.
+Using logic, <math>a+b+c= 10</math>, therefore <math>b+a+1</math>(from the carry over)<math>= 10</math>.
 so b+a=9
 A+1=3

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Difference between revisions of "1991 AJHSME Problems/Problem 20"

Revision as of 23:02, 17 April 2017

Contents

Problem

Solution

Solution

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