Difference between revisions of "2008 AMC 8 Problems/Problem 22"
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\textbf{(E)}\ 34</math> | \textbf{(E)}\ 34</math> | ||
− | ==Solution== | + | ==Solution 1== |
− | + | Instead of finding n, we find <math>x=\frac{n}{3}</math>. We want <math>x</math> and <math>9x</math> to be three-digit whole numbers. The smallest three-digit whole number is <math>100</math>, so that is our minimum value for <math>x</math>, since if <math>x \in \mathbb{Z^+}</math>, then <math>9x \in \mathbb{Z^+}</math>. The largest three-digit whole number divisible by <math>9</math> is <math>999</math>, so our maximum value for <math>x</math> is <math>\frac{999}{9}=111</math>. There are <math>12</math> whole numbers in the closed set <math>\left[100,111\right]</math> , so the answer is <math>\boxed{\textbf{(A)}\ 12}</math>. | |
+ | |||
+ | - ColtsFan10 | ||
==Solution 2== | ==Solution 2== | ||
− | + | ||
+ | We can set the following inequalities up to satisfy the conditions given by the question, | ||
+ | <math>100 \leq \frac{n}{3} \leq 999</math>, | ||
+ | and | ||
+ | <math>100 \leq 3n \leq 999</math>. | ||
+ | Once we simplify these and combine the restrictions, we get the inequality, <math>300 \leq n \leq 333</math>. | ||
+ | Now we have to find all multiples of 3 in this range for <math>\frac{n}{3}</math> to be an integer. We can compute this by setting <math>\frac{n} | ||
+ | {3}=x</math>, where <math>x \in \mathbb{Z^+}</math>. Substituting <math>x</math> for <math>n</math> in the previous inequality, we get, <math>100 \leq x \leq 111</math>, and there are <math>111-100+1</math> integers in this range giving us the answer, <math>\boxed{\textbf{(A)}\ 12}</math>. | ||
+ | |||
+ | - kn07 | ||
+ | |||
+ | ==Solution 3== | ||
+ | |||
+ | We can create a list of the positive integers <math>n</math> that fulfill the requirement of <math>\frac {n}{3}</math> and <math>3n</math> are three-digit whole numbers. The first number of this list must be <math>300</math> since <math>\frac {300}{3} = 100</math> is the smallest positive integer that satisfies this requirement. The last number of this list must be <math>333</math> since <math>3 \cdot 333 = 999</math> is the largest positive integer that satisfies this requirement. Since the problem requires <math>\frac {n}{3}</math> and <math>3n</math> must be whole numbers, the other numbers must be multiples of 3 (just like 300 and 333), so the list would look like this: | ||
+ | <math>300, 303, 306, . . . , 333</math> | ||
+ | To put this list in to a countable form we must put it in a form similar to <math>1,2,3, . . ., n</math>. So, we manipulate it as follows: | ||
+ | <math>300-300,303-300,306-300, . . .,333-300 \Rightarrow 0,3,6, . . ., 33</math> | ||
+ | |||
+ | <math>\frac{0}{3}, \frac{3}{3}, \frac{6}{3}, . . ., \frac{33}{3} \Rightarrow 0,1,2, . . ., 11</math> | ||
+ | |||
+ | <math>0+1,1+1,1+2, . . ., 11+1 \Rightarrow 1,2,3, . . ., 12</math> | ||
+ | |||
+ | Now we can tell that there are 12 positive integers which satisfies the two requirements, so the answer is <math>\boxed{\textbf{(A)}\ 12}</math>. | ||
+ | |||
+ | ~julia333 | ||
+ | |||
+ | |||
− | + | ==Video Solution by OmegaLearn== | |
+ | https://youtu.be/rQUwNC0gqdg?t=230 | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2008|num-b=21|num-a=23}} | {{AMC8 box|year=2008|num-b=21|num-a=23}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 14:34, 30 August 2024
Problem
For how many positive integer values of are both
and
three-digit whole numbers?
Solution 1
Instead of finding n, we find . We want
and
to be three-digit whole numbers. The smallest three-digit whole number is
, so that is our minimum value for
, since if
, then
. The largest three-digit whole number divisible by
is
, so our maximum value for
is
. There are
whole numbers in the closed set
, so the answer is
.
- ColtsFan10
Solution 2
We can set the following inequalities up to satisfy the conditions given by the question,
,
and
.
Once we simplify these and combine the restrictions, we get the inequality,
.
Now we have to find all multiples of 3 in this range for
to be an integer. We can compute this by setting
, where
. Substituting
for
in the previous inequality, we get,
, and there are
integers in this range giving us the answer,
.
- kn07
Solution 3
We can create a list of the positive integers that fulfill the requirement of
and
are three-digit whole numbers. The first number of this list must be
since
is the smallest positive integer that satisfies this requirement. The last number of this list must be
since
is the largest positive integer that satisfies this requirement. Since the problem requires
and
must be whole numbers, the other numbers must be multiples of 3 (just like 300 and 333), so the list would look like this:
To put this list in to a countable form we must put it in a form similar to . So, we manipulate it as follows:
Now we can tell that there are 12 positive integers which satisfies the two requirements, so the answer is .
~julia333
Video Solution by OmegaLearn
https://youtu.be/rQUwNC0gqdg?t=230
See Also
2008 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
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 AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.