Difference between revisions of "2009 UNCO Math Contest II Problems/Problem 2"
m (moved 2009 UNC Math Contest II Problems/Problem 2 to 2009 UNCO Math Contest II Problems/Problem 2: disambiguation of University of Northern Colorado with University of North Carolina) |
Bluelinfish (talk | contribs) m (→Solution) |
||
(4 intermediate revisions by 3 users not shown) | |||
Line 8: | Line 8: | ||
== Solution == | == Solution == | ||
+ | (a) Looking at the units digits, we need the units digit of <math>Q_n</math> to be either <math>0</math> or <math>5</math>. We know that <math>1^n</math> will always have a units digit of <math>1</math>. Looking at <math>2^n</math>, however, cycles every four powers with units digits <math>2, 4, 8,</math> and <math>6</math> in that order. We see that we can only get a units digit of <math>5</math> if we have <math>4</math> as a units digit for <math>2^n</math>, and there is no way to get <math>0</math> as a units digit. Therefore, our answer is <math>\boxed{25}</math> because the four units digits cycle <math>25</math> times in the integers <math>1</math> to <math>100</math>. | ||
+ | |||
+ | |||
+ | (b) Similarly, <math>3^n</math> cycles every four powers with units digits <math>3, 9, 7,</math> and <math>1</math> in that order. And <math>4^n</math> cycles every two powers with units digits <math>4</math> and <math>6</math>. Together the units digit of their sum is <math>0</math> for <math>n=1,2,3 \pmod 4</math> , and <math>4</math> for <math>N=0 \pmod 4</math>. So the answer is <math>\boxed{75}</math>. | ||
== See also == | == See also == | ||
− | {{ | + | {{UNCO Math Contest box|year=2009|n=II|num-b=1|num-a=3}} |
[[Category:Introductory Number Theory Problems]] | [[Category:Introductory Number Theory Problems]] |
Latest revision as of 15:34, 28 May 2021
Problem
(a) Let . For how many
between
and
inclusive is
a multiple of
?
(b) For how many between
and
inclusive is
a multiple of 5?
Solution
(a) Looking at the units digits, we need the units digit of to be either
or
. We know that
will always have a units digit of
. Looking at
, however, cycles every four powers with units digits
and
in that order. We see that we can only get a units digit of
if we have
as a units digit for
, and there is no way to get
as a units digit. Therefore, our answer is
because the four units digits cycle
times in the integers
to
.
(b) Similarly, cycles every four powers with units digits
and
in that order. And
cycles every two powers with units digits
and
. Together the units digit of their sum is
for
, and
for
. So the answer is
.
See also
2009 UNCO Math 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 UNCO Math Contest Problems and Solutions |