Difference between revisions of "1964 AHSME Problems/Problem 16"
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<math>f(x)</math> = <math>x^2</math> + <math>3x</math> + <math>2</math> is congruent to <math>0</math> modulo 6 this has 4 cases, | <math>f(x)</math> = <math>x^2</math> + <math>3x</math> + <math>2</math> is congruent to <math>0</math> modulo 6 this has 4 cases, | ||
When, | When, | ||
− | * <math>x+1</math> | + | * <math>x+1</math> ≡ 0(mod 6) |
− | * <math>x+2</math> | + | * <math>x+2</math> ≡ 0(mod 6) |
− | * <math>x+1</math> | + | * <math>x+1</math> ≡ 2 (mod 6), <math>x+2</math> ≡ 3 (mod 6) |
− | * <math>x+1</math> | + | * <math>x+1</math> ≡ 0 (mod 6), then <math>x+2</math> ≡ 4 (mod 6), which satisfies. |
Therefore by solvin' these cases we get 17. | Therefore by solvin' these cases we get 17. | ||
− | <math> | + | <math>EDITED</math> <math>by</math> <math>RNVAA</math> |
==See Also== | ==See Also== |
Revision as of 11:55, 30 August 2024
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 know that,
=
+
+
is
congruent modulo 6 that implies
or/and
is
congruent modulo
.The numbers are of the form
and
for some integer
and due to restrictions of number of elements in the set
we get the inequality
.Then we calculate the possible combinations to get
as answer i.e. option
.
THE above solution does not give you the answer there are more cases,
=
+
+
is congruent to
modulo 6 this has 4 cases,
When,
≡ 0(mod 6)
≡ 0(mod 6)
≡ 2 (mod 6),
≡ 3 (mod 6)
≡ 0 (mod 6), then
≡ 4 (mod 6), which satisfies.
Therefore by solvin' these cases we get 17.
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 |
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