Difference between revisions of "2003 AMC 12B Problems/Problem 12"
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| − | == Solution == | + | == Solution 1== |
| − | For all consecutive odd integers, one of every five is a multiple of 5 and one of every three is a multiple of 3. The answer is <math>3 \cdot 5 = 15</math>, so <math>{\boxed{\textbf{(D)}}</math> is the correct answer. | + | For all consecutive odd integers, one of every five is a multiple of 5 and one of every three is a multiple of 3. The answer is <math>3 \cdot 5 = 15</math>, so <math>{\boxed{\textbf{(D)15}}}</math> is the correct answer. |
| + | |||
| + | == Solution 2 == | ||
| + | We'll just test all the answer choices. | ||
| + | |||
| + | Note that for any 3 consecutive odd integers, there will be exactly one multiple of <math>3.</math> | ||
| + | |||
| + | Let's list all possibilities of 3 consecutive odd integers. (multiple of 3, not multiple of 3, not multiple of 3), (not multiple of 3, multiple of 3, not multiple of 3) and (not multiple of 3, not multiple of 3, multiple of 3) | ||
| + | |||
| + | To support this further, list the first few consecutive lists of 3 consecutive odd integers. | ||
| + | |||
| + | We have <math>(1, 3, 5), (3, 5, 7), (5, 7, 9), (7, 9, 11), (11, 13, 15), \ldots</math> | ||
| + | |||
| + | So if the first two consecutive odd numbers aren't multiples of three, the last one must be a multiple of three. | ||
| + | |||
| + | Therefore for five consecutive odd integers, there must be at least one multiple of three. | ||
| + | |||
| + | In the same fashion as we did above, note that for any 5 consecutive integers, there will also be exactly one multiple of <math>5.</math> | ||
| + | |||
| + | Therefore, for any 5 consecutive odd integers, there must be exactly one multiple of five. | ||
| + | |||
| + | We can skip 7 since none of the answer choices are a multiple of 7. | ||
| + | |||
| + | Now we try <math>11.</math> 11 doesn't work since as we see the first set of 5 consecutive odd integers doesn't fit, namely <math>(1, 3, 5, 7, 9).</math> | ||
| + | |||
| + | Since any 5 consecutive integers is divisible both by <math>3</math> and <math>5</math>, it also must be divisible by <math>{\boxed{\textbf{(D)15}}}</math> and no higher since we saw that <math>11</math> does not work and that there is no answer choice that is multiple of <math>7</math>. | ||
| + | |||
| + | ~mathboy282 | ||
| + | |||
| + | ==Video Solution by WhyMath== | ||
| + | https://youtu.be/dQiw5lisd84 | ||
| + | |||
| + | ~savannahsolver | ||
==See Also== | ==See Also== | ||
Latest revision as of 15:43, 29 June 2021
- The following problem is from both the 2003 AMC 12B #12 and 2003 AMC 10B #18, so both problems redirect to this page.
Problem
What is the largest integer that is a divisor of
![\[(n+1)(n+3)(n+5)(n+7)(n+9)\]](http://latex.artofproblemsolving.com/c/6/9/c694dd7f6739034188fe758646f1fd28508df784.png)
for all positive even integers 
?
Solution 1
For all consecutive odd integers, one of every five is a multiple of 5 and one of every three is a multiple of 3. The answer is 
, so 
is the correct answer.
Solution 2
We'll just test all the answer choices.
Note that for any 3 consecutive odd integers, there will be exactly one multiple of 
Let's list all possibilities of 3 consecutive odd integers. (multiple of 3, not multiple of 3, not multiple of 3), (not multiple of 3, multiple of 3, not multiple of 3) and (not multiple of 3, not multiple of 3, multiple of 3)
To support this further, list the first few consecutive lists of 3 consecutive odd integers.
We have 
So if the first two consecutive odd numbers aren't multiples of three, the last one must be a multiple of three.
Therefore for five consecutive odd integers, there must be at least one multiple of three.
In the same fashion as we did above, note that for any 5 consecutive integers, there will also be exactly one multiple of 
Therefore, for any 5 consecutive odd integers, there must be exactly one multiple of five.
We can skip 7 since none of the answer choices are a multiple of 7.
Now we try 
11 doesn't work since as we see the first set of 5 consecutive odd integers doesn't fit, namely 
Since any 5 consecutive integers is divisible both by 
and 
, it also must be divisible by and no higher since we saw that 
does not work and that there is no answer choice that is multiple of 
.
~mathboy282
Video Solution by WhyMath
~savannahsolver
See Also
| 2003 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 11 |
Followed by Problem 13 |
| 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 | |
| 2003 AMC 10B (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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
