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Difference between revisions of "1967 AHSME Problems/Problem 1"

(Created page with "==Problem== The three-digit number <math>2a3</math> is added to the number <math>326</math> to give the three-digit number <math>5b9</math>. If <math>5b9</math> is divisible by...")
 
(→‎Solution)
Line 7: Line 7:
 
==Solution==
 
==Solution==
  
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If <math>5b9</math> is divisible by <math>9</math>, this must mean that <math>5 + b + 9</math> is a multiple of <math>9</math>. So, <math>5 + b + 9 = 9, 18, 27, 36...</math>. Because <math>5 + 9 = 14</math> and <math>b</math> is in between 0 and 9, <math>5 + b + 9 = 18</math> and <math>b = 4</math>.
+
If <math>5b9</math> is divisible by <math>9</math>, this must mean that <math>5 + b + 9</math> is a multiple of <math>9</math>. So, <math>5 + b + 9 = 9, 18, 27, 36...</math>.  
  
−
<math>2a3 + 326 = 549</math>, so <math>2a3 = 549 - 326</math> and <math>a = 2</math>
+
Because <math>5 + 9 = 14</math> and <math>b</math> is in between 0 and 9,  
 +
<math></math>5 + b + 9 = 18<math> and </math>b = 4<math></math>
  
−
<math>a + b = 6</math> which is answer choice <math>\boxed{C}</math>.
+
<math>2a3 + 326 = 549</math>, so
 +
<cmath>2a3 = 549 - 326</cmath>
 +
<cmath>a = 2</cmath>
 +
 
 +
<cmath>a + b = 6</cmath>
 +
which is answer choice <math>\boxed{C}</math>.
  
 
==See Also==
 
==See Also==

Revision as of 21:35, 5 June 2012

Problem

The three-digit number $2a3$ is added to the number $326$ to give the three-digit number $5b9$. If $5b9$ is divisible by 9, then $a+b$ equals

$\text{(A)}\ 2\qquad\text{(B)}\ 4\qquad\text{(C)}\ 6\qquad\text{(D)}\ 8\qquad\text{(E)}\ 9$

Solution

If $5b9$ is divisible by $9$, this must mean that $5 + b + 9$ is a multiple of $9$. So, $5 + b + 9 = 9, 18, 27, 36...$.

Because $5 + 9 = 14$ and $b$ is in between 0 and 9, $$ (Error compiling LaTeX. Unknown error_msg)5 + b + 9 = 18$and$b = 4$$ (Error compiling LaTeX. Unknown error_msg)

$2a3 + 326 = 549$, so \[2a3 = 549 - 326\] \[a = 2\]

\[a + b = 6\] which is answer choice $\boxed{C}$.

See Also

1967 AHSC (Problems • Answer Key • Resources)
Preceded by
First Question 		Followed by
Problem 2
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All AHSME Problems and Solutions
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