Difference between revisions of "2018 UNM-PNM Statewide High School Mathematics Contest II Problems/Problem 2"
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Sub case 2: k is congruent to -3(mod 8) | Sub case 2: k is congruent to -3(mod 8) | ||
− | <math>{(-3)^2}</math> + <math>-3</math> + <math>3</math> = 1(mod 8) | + | <math>{(-3)^2}</math> + (<math>-3</math>) + <math>3</math> = 1(mod 8) |
Sub case 3: k is congruent to -2(mod 8) | Sub case 3: k is congruent to -2(mod 8) | ||
− | <math>{(-2)^2}</math> + <math>-2</math> + <math>3</math> = 5(mod 8) | + | <math>{(-2)^2}</math> + (<math>-2</math>) + <math>3</math> = 5(mod 8) |
Sub case 4: k is congruent to -1(mod 8) | Sub case 4: k is congruent to -1(mod 8) | ||
− | <math>{(-1)^2}</math> + <math>-1</math> + <math>3</math> = 3(mod 8) | + | <math>{(-1)^2}</math> + (<math>-1</math>) + <math>3</math> = 3(mod 8) |
Sub case 5: k is congruent to 0(mod 8) | Sub case 5: k is congruent to 0(mod 8) | ||
− | <math>{( | + | <math>{(0)^2}</math> + <math>0</math> + <math>3</math> = 3(mod 8) |
Sub case 6: k is congruent to 1(mod 8) | Sub case 6: k is congruent to 1(mod 8) |
Revision as of 17:42, 21 August 2023
Problem
Determine all positive integers such that
and
is divisible by
.
Solution
If is divisible by
, then so is
.
is the same as
, and it is divisible by
. There are many options:
is divisible by
,
is divisible by
,
is divisible by
and
is divisible by
, and
is divisible by
and
is divisible by
.
Case 1: is divisible by
This means that is congruent to
. Satisfying the range, the following integers that satisfy are:
{}
Case 2: is divisible by
Or is in the form
. This means that
can be {
}, in the list, the first three numbers are prime, and The fourth can be factorized into two non-consecutive primes. No results from this case.
Case 3: is divisible by
and
is divisible by
is congruent to
or
=
+
where
is a positive integer. This means that
+
+
=
+
+
which has to be divisible by
. That means so does
, or
itself is divisible by
. The maximum it can be
, because
or
=
. However, for the available values that can be inputted (0,3,6,9,and 12),the same list results from Case 1. No new values.
Case 4: is divisible by
and
is divisible by
is congruent to
or
=
+
where
is a positive integer. This means that
+
+
=
+
+
which has to be divisible by
. That means so does
+
+
. Checking k modulo eight for all values might result in a value of k which can narrow down search values.
Sub case 1: k is congruent to -4(mod 8)
+
+
= 7(mod 8)
Sub case 2: k is congruent to -3(mod 8)
+ (
) +
= 1(mod 8)
Sub case 3: k is congruent to -2(mod 8)
+ (
) +
= 5(mod 8)
Sub case 4: k is congruent to -1(mod 8)
+ (
) +
= 3(mod 8)
Sub case 5: k is congruent to 0(mod 8)
+
+
= 3(mod 8)
Sub case 6: k is congruent to 1(mod 8)
+
+
= 5(mod 8)
Sub case 7: k is congruent to 2(mod 8)
+
+
= 1(mod 8)
Sub case 8: k is congruent to 3(mod 8)
+
+
= 7(mod 8)
In no scenario is +
+
divisible by
, and by working backward, neither can
. This means that the list noted in Case 1 are all the numbers possible that satisfy the condition. Our answer is
See also
2018 UNM-PNM Contest II (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 | ||
All UNM-PNM Problems and Solutions |