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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...")
 
m (→‎See Also)
 
(7 intermediate revisions by 2 users not shown)
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==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>.
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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, <cmath>5 + b + 9 = 9, 18, 27, 36...</cmath>  
  
−
<math>2a3 + 326 = 549</math>, so <math>2a3 = 549 - 326</math> and <math>a = 2</math>
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Because <math>5 + 9 = 14</math> and <math>b</math> is in between 0 and 9,
  
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<math>a + b = 6</math> which is answer choice <math>\boxed{C}</math>.
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<cmath>5 + b + 9 = 18</cmath>
 +
<cmath>b = 4</cmath>
 +
 
 +
The question states that
 +
<cmath>2a3 + 326 = 549</cmath>
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so
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<cmath>2a3 = 549 - 326</cmath>
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<cmath>2a3 = 223</cmath>
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<cmath>a = 2</cmath>
 +
 
 +
<cmath>a + b = 6</cmath>
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which is answer choice <math>\boxed{C}</math>.
  
 
==See Also==
 
==See Also==
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{{AHSME 50p box|year=1967|before=First Question|num-a=2}}
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{{AHSME 40p box|year=1967|before=First Question|num-a=2}}
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 00:45, 16 August 2023

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,

\[5 + b + 9 = 18\] \[b = 4\]

The question states that \[2a3 + 326 = 549\] so \[2a3 = 549 - 326\] \[2a3 = 223\] \[a = 2\]

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

See Also

1967 AHSC (Problems • Answer Key • Resources)
Preceded by
First Question 		Followed by
Problem 2
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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
All AHSME Problems and Solutions

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