Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1951 AHSME Problems/Problem 18"

m (typo in problem)
m (→‎Problem)
 
(7 intermediate revisions by 2 users not shown)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
−
The expression <math> 21x^2 +ax +21</math> is to be factored into two linear prime binomial factors with integer coefficients. This can be done if <math> a</math> is:
+
The expression <math>21x^2+ax+21</math> is to be factored into prime binomial factors and without a numerical monomial factor. This can be done if the value ascribed to <math>a</math> is:
  
 
<math> \textbf{(A)}\ \text{any odd number} \qquad\textbf{(B)}\ \text{some odd number} \qquad\textbf{(C)}\ \text{any even number}</math>
 
<math> \textbf{(A)}\ \text{any odd number} \qquad\textbf{(B)}\ \text{some odd number} \qquad\textbf{(C)}\ \text{any even number}</math>
Line 6: Line 6:
  
 
== Solution ==
 
== Solution ==
−
We can factor <math> 21x^2 + ax + 21</math> as <math>(7x+3)(3x+7)</math>, which expands to <math>21x^2+42x+21</math>. So the answer is <math> \textbf{(D)}\ \text{some even number}</math>
+
We can factor <math> 21x^2 + ax + 21</math> as <math>(7x+3)(3x+7)</math>, which expands to <math>21x^2+58x+21</math>. So the answer is <math> \textbf{(D)}\ \text{some even number}</math>
 +
 
 +
== Solution 2==
 +
Factoring <math>21x^2+ax+21</math> by grouping, we need to find some <math>b,c</math> such that <math>b\cdot c = 21^2</math>, and that <math>b+c=a</math>.
 +
Since <math>21^2\equiv 1\;(mod\;2)</math>, <math>b\land c \equiv 1\;(mod\;2)</math>, and <math>b+c \equiv 0\;(mod\;2)</math>. So <math>a</math> must be even. <math>a</math> cannot be <math>any</math> even number, since <math>21^2</math> only has 4 odd factors, so the answer is <math> \textbf{(D)}\ \text{some even number}</math>
  
 
== See Also ==
 
== See Also ==

Latest revision as of 12:29, 6 June 2024

Problem

The expression $21x^2+ax+21$ is to be factored into prime binomial factors and without a numerical monomial factor. This can be done if the value ascribed to $a$ is:

$\textbf{(A)}\ \text{any odd number} \qquad\textbf{(B)}\ \text{some odd number} \qquad\textbf{(C)}\ \text{any even number}$ $\textbf{(D)}\ \text{some even number} \qquad\textbf{(E)}\ \text{zero}$

Solution

We can factor $21x^2 + ax + 21$ as $(7x+3)(3x+7)$, which expands to $21x^2+58x+21$. So the answer is $\textbf{(D)}\ \text{some even number}$

Solution 2

Factoring $21x^2+ax+21$ by grouping, we need to find some $b,c$ such that $b\cdot c = 21^2$, and that $b+c=a$. Since $21^2\equiv 1\;(mod\;2)$, $b\land c \equiv 1\;(mod\;2)$, and $b+c \equiv 0\;(mod\;2)$. So $a$ must be even. $a$ cannot be $any$ even number, since $21^2$ only has 4 odd factors, so the answer is $\textbf{(D)}\ \text{some even number}$

See Also

1951 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
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 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1951_AHSME_Problems/Problem_18&oldid=220149"