Difference between revisions of "2001 AIME I Problems/Problem 1"
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Find the sum of all positive two-digit integers that are divisible by each of their digits. | Find the sum of all positive two-digit integers that are divisible by each of their digits. | ||
− | == Solution == | + | == Solution 1 == |
Let our number be <math>10a + b</math>, <math>a,b \neq 0</math>. Then we have two conditions: <math>10a + b \equiv 10a \equiv 0 \pmod{b}</math> and <math>10a + b \equiv b \pmod{a}</math>, or <math>a</math> divides into <math>b</math> and <math>b</math> divides into <math>10a</math>. Thus <math>b = a, 2a,</math> or <math>5a</math> (note that if <math>b = 10a</math>, then <math>b</math> would not be a digit). | Let our number be <math>10a + b</math>, <math>a,b \neq 0</math>. Then we have two conditions: <math>10a + b \equiv 10a \equiv 0 \pmod{b}</math> and <math>10a + b \equiv b \pmod{a}</math>, or <math>a</math> divides into <math>b</math> and <math>b</math> divides into <math>10a</math>. Thus <math>b = a, 2a,</math> or <math>5a</math> (note that if <math>b = 10a</math>, then <math>b</math> would not be a digit). | ||
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If we ignore the case <math>b = 0</math> as we have been doing so far, then the sum is <math>495 + 120 + 15 = \boxed{630}</math>. | If we ignore the case <math>b = 0</math> as we have been doing so far, then the sum is <math>495 + 120 + 15 = \boxed{630}</math>. | ||
− | Using casework, we can list out all of these numbers: < | + | == Solution 2 == |
+ | |||
+ | Using casework, we can list out all of these numbers: <cmath>11+12+15+22+24+33+36+44+48+55+66+77+88+99=\boxed{630}.</cmath> | ||
+ | |||
+ | == Solution 3 == | ||
+ | |||
+ | To further expand on solution 2, it would be tedious to test all <math>90</math> two-digit numbers. We can reduce the amount to look at by focusing on the tens digit. | ||
+ | First, we cannot have any number that is a multiple of <math>10</math>. We also note that any number with the same digits is a number that satisfies this problem. This gives <cmath>11, 22, 33, ... 99.</cmath> We start from each of these numbers and constantly add the digit of the tens number of the respective number until we get a different tens digit. For example, we look at numbers <math>11, 12, 13, ... 19</math> and numbers <math>22, 24, 26, 28</math>. This heavily reduces the numbers we need to check, as we can deduce that any number with a tens digit of <math>5</math> or greater that does not have two of the same digits is not a valid number for this problem. This will give us the numbers from solution 2. | ||
+ | |||
+ | == Solution 4 == | ||
+ | |||
+ | In this solution, we will do casework on the ones digit. | ||
+ | Before we start, let's make some variables. Let <math>a</math> be the ones digit, and <math>b</math> be the tens digit. Let <math>n</math> equal our number. Our number can be expressed as <math>10b+a</math>. We can easily see that <math>b|a</math>, since <math>b|n</math>, and <math>b|10b</math>. Therefore, <math>b|(n-10b)</math>. | ||
+ | Now, let's start with the casework. | ||
+ | |||
+ | Case 1: <math>a=1</math> | ||
+ | Since <math>b|a</math>, <math>b=1</math>. From this, we get that <math>n=11</math> satisfies the condition. | ||
+ | |||
+ | Case 2: <math>a=2</math> | ||
+ | We either have <math>b=1</math>, or <math>b=2</math>. From this, we get that <math>n=12</math> and <math>n=22</math> satisfy the condition. | ||
+ | |||
+ | Case 3: <math>a=3</math> | ||
+ | We have <math>b=3</math>. From this, we get that <math>n=33</math> satisfies the condition. Note that <math>b=1</math> was not included because <math>3</math> does not divide <math>13</math>. | ||
+ | |||
+ | Case 4: <math>a=4</math> | ||
+ | We either have <math>b=2</math> or <math>b=4</math>. From this, we get that <math>n=24</math> and <math>n=44</math> satisfy the condition. <math>b=1</math> was not included for similar reasons as last time. | ||
+ | |||
+ | Case 5: <math>a=5</math> | ||
+ | We either have <math>b=1</math> or <math>b=5</math>. From this, we get that <math>n=15</math> and <math>n=55</math> satisfy the condition. | ||
+ | |||
+ | Continuing with this process up to <math>a=9</math>, we get that <math>n</math> could be <math>11, 12, 22, 33, 24, 44, 15, 55, 36, 66, 77, 48, 88, 99</math>. Summing, we get that the answer is <math>\boxed{630}</math>. A clever way to sum would be to group the multiples of <math>11</math> together to get <math>11+22+\dots+99=(45)(11)=495</math>, and then add the remaining <math>12+24+15+36+48=135</math>. | ||
+ | |||
+ | <i>-bronzetruck2016</i> | ||
== See also == | == See also == |
Latest revision as of 18:37, 27 November 2022
Problem
Find the sum of all positive two-digit integers that are divisible by each of their digits.
Solution 1
Let our number be ,
. Then we have two conditions:
and
, or
divides into
and
divides into
. Thus
or
(note that if
, then
would not be a digit).
- For
, we have
for nine possibilities, giving us a sum of
.
- For
, we have
for four possibilities (the higher ones give
), giving us a sum of
.
- For
, we have
for one possibility (again, higher ones give
), giving us a sum of
.
If we ignore the case as we have been doing so far, then the sum is
.
Solution 2
Using casework, we can list out all of these numbers:
Solution 3
To further expand on solution 2, it would be tedious to test all two-digit numbers. We can reduce the amount to look at by focusing on the tens digit.
First, we cannot have any number that is a multiple of
. We also note that any number with the same digits is a number that satisfies this problem. This gives
We start from each of these numbers and constantly add the digit of the tens number of the respective number until we get a different tens digit. For example, we look at numbers
and numbers
. This heavily reduces the numbers we need to check, as we can deduce that any number with a tens digit of
or greater that does not have two of the same digits is not a valid number for this problem. This will give us the numbers from solution 2.
Solution 4
In this solution, we will do casework on the ones digit.
Before we start, let's make some variables. Let be the ones digit, and
be the tens digit. Let
equal our number. Our number can be expressed as
. We can easily see that
, since
, and
. Therefore,
.
Now, let's start with the casework.
Case 1:
Since
,
. From this, we get that
satisfies the condition.
Case 2:
We either have
, or
. From this, we get that
and
satisfy the condition.
Case 3:
We have
. From this, we get that
satisfies the condition. Note that
was not included because
does not divide
.
Case 4:
We either have
or
. From this, we get that
and
satisfy the condition.
was not included for similar reasons as last time.
Case 5:
We either have
or
. From this, we get that
and
satisfy the condition.
Continuing with this process up to , we get that
could be
. Summing, we get that the answer is
. A clever way to sum would be to group the multiples of
together to get
, and then add the remaining
.
-bronzetruck2016
See also
2001 AIME I (Problems • Answer Key • Resources) | ||
Preceded by First Question |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.