Difference between revisions of "1964 AHSME Problems/Problem 16"
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− | We have | + | We have f(<math>x</math>)=<math>x</math>^<math>2</math> + <math>3</math><math>x</math> +<math>2</math> and <math>S</math> = {<math>0</math>,<math>1</math>,<math>2</math>,... ,<math>25</math>}.We need equiv <math>f</math>(<math>s</math>)=0 mod 6<math>. That implies we need numbers of the form 6</math>k<math> +</math>5<math> and 6</math>k<math> +</math>4<math> also due to restriction on number of elements of set we need </math>k<math><= 3 . Calculate the number of possibble combinations to get </math>17<math> as answer.</math>Ans.<math> </math>\boxed{\textbf{(E)}}<math>. |
Solution by </math>GEOMETRY-WIZARD$. | Solution by </math>GEOMETRY-WIZARD$. |
Revision as of 03:24, 31 December 2023
Contents
Problem
Let and let
be the set of integers
.
The number of members
of
such that
has remainder zero when divided by
is:
Solution ![$1$](//latex.artofproblemsolving.com/d/c/e/dce34f4dfb2406144304ad0d6106c5382ddd1446.png)
Note that for all polynomials ,
.
Proof:
If , then
. In the second equation, we can use the binomial expansion to expand every term, and then subtract off all terms that have a factor of
or higher, since subtracting a multiple of
will not change congruence
. This leaves
, which is
, so
.
So, we only need to test when has a remainder of
for
. The set of numbers
will repeat remainders, as will all other sets. The remainders are
.
This means for ,
is divisible by
. Since
is divisible, so is
for
, which is
values of
that work. Since
is divisible, so is
for
, which is
more values of
that work. The values of
will also generate
solutions each, just like
. This is a total of
values of
, for an answer of
Solution 2
We have f()=
^
+
+
and
= {
,
,
,... ,
}.We need equiv
(
)=0 mod 6
k
5
k
4
k
17
Ans.$$ (Error compiling LaTeX. Unknown error_msg)\boxed{\textbf{(E)}}$.
Solution by$ (Error compiling LaTeX. Unknown error_msg)GEOMETRY-WIZARD$.
See Also
1964 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
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.