Difference between revisions of "2024 AMC 10A Problems/Problem 18"
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\lfloor(2024-3)/8\rfloor+\lfloor(2024-6)/8\rfloor+\lfloor(2024-7)/8\rfloor+3=759</math> | \lfloor(2024-3)/8\rfloor+\lfloor(2024-6)/8\rfloor+\lfloor(2024-7)/8\rfloor+3=759</math> | ||
take away one because <math>b=3</math> is out of range, so <math>758\Rightarrow7+8+5=\boxed{\text{(D) }20}</math> | take away one because <math>b=3</math> is out of range, so <math>758\Rightarrow7+8+5=\boxed{\text{(D) }20}</math> | ||
+ | |||
+ | == Video Solution by Pi Academy == | ||
+ | |||
+ | https://youtu.be/fW7OGWee31c?si=oq7toGPh2QaksLHE | ||
+ | |||
==See also== | ==See also== |
Revision as of 13:56, 9 November 2024
- The following problem is from both the 2024 AMC 10A #18 and 2024 AMC 12A #11, so both problems redirect to this page.
Contents
[hide]Problem
There are exactly positive integers
with
such that the base-
integer
is divisible by
(where
is in base ten). What is the sum of the digits of
?
Solution 1
, if
even then
. If
odd then
so
. Now
so
but
is too small so
.
~OronSH ~mathkiddus ~andliu766
Solution 2
Clearly, is either even or odd. If
is even, let
.
Thus, one solution is for some integer
, or
.
What if is odd? Then let
:
This simply states that is odd. Thus, the other solution is
for some integer
, or
.
We now simply must count the number of integers between and
, inclusive, that are
mod
or
mod
. Note that the former case comprises even numbers only while the latter is only odd; thus, there is no overlap and we can safely count the number of each and add them.
In the former case, we have the numbers ; this list is equivalent to
, which comprises
numbers. In the latter case, we have the numbers
, which comprises
numbers. There are
numbers in total, so our answer is
.
~Technodoggo
Solution 3
Note that is to be divisible by
, which means that
is divisible by
.
If , then
is not divisible by
.
If , then
is not divisible by
.
If , then
is not divisible by
.
If , then
is divisible by
.
If , then
is not divisible by
.
If , then
is not divisible by
.
If , then
is divisible by
.
If , then
is divisible by
.
Therefore, for every values of
,
of them will make
divisible by
. Therefore, since
is divisible by
,
values of
, but this includes
, which does not satisfy the given inequality. Therefore, the answer is
~Tacos_are_yummy_1
Solution 4
We need
take away one because
is out of range, so
Video Solution by Pi Academy
https://youtu.be/fW7OGWee31c?si=oq7toGPh2QaksLHE
See also
2024 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
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 | ||
All AMC 10 Problems and Solutions |
2024 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 10 |
Followed by Problem 12 |
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 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.