Difference between revisions of "1965 AHSME Problems/Problem 19"
m (added letter choice to final answers) |
m (fixed typos that made it not work) |
||
Line 60: | Line 60: | ||
== See Also == | == See Also == | ||
− | {{AHSME 40p box|year=1965|num-b= | + | {{AHSME 40p box|year=1965|num-b=18|num-a=20}} |
{{MAA Notice}} | {{MAA Notice}} | ||
[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] |
Revision as of 16:57, 18 July 2024
Contents
Problem 19
If is exactly divisible by
, the value of
is:
Solution 1
Let and
.
Let 3 roots of be
and
. As
, 3 roots of 4 roots of
will be same as roots of
. Let the 4th root of
be
. By vieta's formula
In
In
so
By ~Ahmed_Ashhab
Solution 2
Notice that to obtain the term one must multiply
by some linear function of the form
. Looking at the
term, it is clear that
must equal
. Therefore by multiplying
by
, the product will be
. Therefore
,
,
. Thus
See Also
1965 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.