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

Latest revision as of 00:45, 16 August 2023

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

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

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 $\boxed{C}$ .

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

@@ Line 7: / Line 7: @@
 ==Solution==
-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>.
+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>
 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>
+<cmath>5 + b + 9 = 18</cmath>
+<cmath>b = 4</cmath>
-<math>2a3 + 326 = 549</math>, so
+The question states that
+<cmath>2a3 + 326 = 549</cmath>
+so
 <cmath>2a3 = 549 - 326</cmath>
+<cmath>2a3 = 223</cmath>
 <cmath>a = 2</cmath>
@@ Line 21: / Line 25: @@
 ==See Also==
-{{AHSME 50p box|year=1967|before=First Question|num-a=2}}
+{{AHSME 40p box|year=1967|before=First Question|num-a=2}}
 [[Category:Introductory Algebra Problems]]
+{{MAA Notice}}