Difference between revisions of "2021 AMC 10B Problems/Problem 16"
MRENTHUSIASM (talk | contribs) m (Added in the "See Also" and MAA Notice.) |
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<math>\textbf{(A)} ~4 \qquad\textbf{(B)} ~5 \qquad\textbf{(C)} ~6 \qquad\textbf{(D)} ~7 \qquad\textbf{(E)} ~8</math> | <math>\textbf{(A)} ~4 \qquad\textbf{(B)} ~5 \qquad\textbf{(C)} ~6 \qquad\textbf{(D)} ~7 \qquad\textbf{(E)} ~8</math> | ||
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==Solution 1== | ==Solution 1== | ||
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The divisibility rule of <math>15</math> is that the number must be congruent to <math>0</math> mod <math>3</math> and congruent to <math>0</math> mod <math>5</math>. Being divisible by <math>5</math> means that it must end with a <math>5</math> or a <math>0</math>. We can rule out the case when the number ends with a <math>0</math> immediately because the only integer that is uphill and ends with a <math>0</math> is <math>0</math> which is not positive. So now we know that the number ends with a <math>5</math>. 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 <math>3</math>. These numbers are <math>15, 45, 135, 345, 1245, 12345</math> which are <math>6</math> numbers C. | The divisibility rule of <math>15</math> is that the number must be congruent to <math>0</math> mod <math>3</math> and congruent to <math>0</math> mod <math>5</math>. Being divisible by <math>5</math> means that it must end with a <math>5</math> or a <math>0</math>. We can rule out the case when the number ends with a <math>0</math> immediately because the only integer that is uphill and ends with a <math>0</math> is <math>0</math> which is not positive. So now we know that the number ends with a <math>5</math>. 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 <math>3</math>. These numbers are <math>15, 45, 135, 345, 1245, 12345</math> which are <math>6</math> numbers C. | ||
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==Solution 2== | ==Solution 2== | ||
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~coolmath34 | ~coolmath34 | ||
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+ | ==Solution 4== | ||
+ | For every positive integer, it is divisible by <math>3</math> if and only if its digit-sum is divisible by <math>3;</math> it is divisible by <math>5</math> if and only if its units digit is <math>0</math> or <math>5.</math> Since the desired positive integers are going <b>uphill</b>, their units digits must be <math>5</math>s. | ||
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+ | <b>Case (1): Remove <math>\boldsymbol{0}</math> digits.</b> | ||
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+ | <b>Case (2): Remove <math>\boldsymbol{1}</math> digit.</b> | ||
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+ | <b>Case (3): Remove <math>\boldsymbol{2}</math> digits.</b> | ||
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+ | <b>Case (4): Remove <math>\boldsymbol{3}</math> digits.</b> | ||
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+ | <b>Case (5): Remove <math>\boldsymbol{4}</math> digits.</b> | ||
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+ | ~MRENTHUSIASM | ||
== Video Solution by OmegaLearn (Using Divisibility Rules and Casework) == | == Video Solution by OmegaLearn (Using Divisibility Rules and Casework) == |
Revision as of 14:17, 5 March 2021
Contents
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
Solution 3
Like solution 2, we can proceed by using casework. A number is divisible by if is divisible by
and
In this case, the units digit must be
otherwise no number can be formed.
Case 1: sum of digits = 6
There is only one number,
Case 2: sum of digits = 9
There are two numbers: and
Case 3: sum of digits = 12
There are two numbers: and
Case 4: sum of digits = 15
There is only one number,
We can see that we have exhausted all cases, because in order to have a larger sum of digits, then a number greater than needs to be used, breaking the conditions of the problem. The answer is
~coolmath34
Solution 4
For every positive integer, it is divisible by if and only if its digit-sum is divisible by
it is divisible by
if and only if its units digit is
or
Since the desired positive integers are going uphill, their units digits must be
s.
Case (1): Remove digits.
Case (2): Remove digit.
Case (3): Remove digits.
Case (4): Remove digits.
Case (5): Remove digits.
~MRENTHUSIASM
Video Solution by OmegaLearn (Using Divisibility Rules and Casework)
~ pi_is_3.14
Video Solution by TheBeautyofMath
~IceMatrix
Video Solution by Interstigation
~Interstigation
See Also
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.