Difference between revisions of "2021 AMC 10B Problems/Problem 16"
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==Solution 2== | ==Solution 2== | ||
First, note how the number must end in either <math>5</math> or <math>0</math> in order to satisfying being divisible by <math>15</math>. However, the number can't end in <math>0</math> because it's not strictly greater than the previous digits. Thus, our number must end in <math>5</math>. We do casework on the number of digits. <math>\newline</math> | First, note how the number must end in either <math>5</math> or <math>0</math> in order to satisfying being divisible by <math>15</math>. However, the number can't end in <math>0</math> because it's not strictly greater than the previous digits. Thus, our number must end in <math>5</math>. We do casework on the number of digits. <math>\newline</math> | ||
− | Case <math> | + | |
− | Case <math> | + | '''Case 1''' = <math>1</math> digit. No numbers work, so <math>0</math> <math>\newline</math> |
− | Case <math> | + | |
− | Case <math> | + | '''Case 2''' = <math>2</math> digits. We have the numbers <math>15, 45,</math> and <math>75</math>, but <math>75</math> isn't an uphill number, so <math>2</math> numbers. <math>\newline</math> |
− | Case <math> | + | |
+ | '''Case 3''' = <math>3</math> digits. We have the numbers <math>135, 345</math>. So <math>2</math> numbers. <math>\newline</math> | ||
+ | |||
+ | '''Case 4''' = <math>4</math> digits. We have the numbers <math>1235, 1245</math> and <math>2345</math>, but only <math>1245</math> satisfies this condition, so <math>1</math> number. <math>\newline</math> | ||
+ | |||
+ | '''Case 5'''= <math>5</math> digits. We have only <math>12345</math>, so <math>1</math> number. <math>\newline</math> | ||
+ | |||
Adding these up, we have <math>2+2+1+1 = 6</math>. <math>\boxed {C}</math> | Adding these up, we have <math>2+2+1+1 = 6</math>. <math>\boxed {C}</math> | ||
Revision as of 03:04, 12 February 2021
Problem
Call a positive integer an uphill integer if every digit is strictly greater than the previous digit. For example, and
are all uphill integers, but
and
are not. How many uphill integers are divisible by
?
Solution 1
The divisibility rule of is that the number must be congruent to
mod
and congruent to
mod
. Being divisible by
means that it must end with a
or a
. We can rule out the case when the number ends with a
immediately because the only integer that is uphill and ends with a
is
which is not positive. So now we know that the number ends with a
. Looking at the answer choices, the answer choices are all pretty small, so we can generate all of the numbers that are uphill and are divisible by
. These numbers are
which are
numbers C.
Solution 2
First, note how the number must end in either or
in order to satisfying being divisible by
. However, the number can't end in
because it's not strictly greater than the previous digits. Thus, our number must end in
. We do casework on the number of digits.
Case 1 = digit. No numbers work, so
Case 2 = digits. We have the numbers
and
, but
isn't an uphill number, so
numbers.
Case 3 = digits. We have the numbers
. So
numbers.
Case 4 = digits. We have the numbers
and
, but only
satisfies this condition, so
number.
Case 5= digits. We have only
, so
number.
Adding these up, we have .
~JustinLee2017
2021 AMC 10B (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 | ||
All AMC 10 Problems and Solutions |