Difference between revisions of "2020 AMC 10A Problems/Problem 22"

Revision as of 22:11, 1 February 2020

Problem

For how many positive integers $n \le 1000$ is $\left\lfloor \dfrac{998}{n} \right\rfloor+\left\lfloor \dfrac{999}{n} \right\rfloor+\left\lfloor \dfrac{1000}{n}\right \rfloor$ not divisible by $3$ ? (Recall that $\lfloor x \rfloor$ is the greatest integer less than or equal to $x$ .)

$\textbf{(A) } 22 \qquad\textbf{(B) } 23 \qquad\textbf{(C) } 24 \qquad\textbf{(D) } 25 \qquad\textbf{(E) } 26$

Solution (Casework)

Expression: $\left\lfloor \dfrac{998}{n} \right\rfloor+\left\lfloor \dfrac{999}{n} \right\rfloor+\left\lfloor \dfrac{1000}{n}\right \rfloor$

Solution:

Let $a = \left\lfloor \frac{998}n \right\rfloor$

Notice that for every integer $n \neq 1$ ,

$\bullet$ if $\frac{998}n$ is an integer, then the three terms in the expression above must be $(a, a, a)$ ,

$\bullet$ if $\frac{999}n$ is an integer, then the three terms in the expression above must be $(a, a + 1, a + 1)$ , and

$\bullet$ if $\frac{1000}n$ is an integer, then the three terms in the expression above must be $(a, a, a + 1)$ .

$\bullet$ if none of { $\frac{998}n$ , $\frac{999}n$ , $\frac{1000}n$ } are integral, then the three terms in the expression above must be $(a, a, a)$ .

This is due to the fact that $998$ , $999$ , and $1000$ do not collectively share any common factors (other than 1).

Note that $n = 1$ doesn't work; to prove this, we just have to substitute $1$ for $n$ in the expression. This gives us $\left\lfloor \dfrac{998}{1} \right\rfloor+\left\lfloor \dfrac{999}{1} \right\rfloor+\left\lfloor \dfrac{1000}{1}\right \rfloor = \frac{998}1 + \frac{999}1 + \frac{1000}1 = 2997 = 999 \cdot 3$ which is divisible by 3. Therefore, the case $n = 1$ does not work.

Now, we test the four cases listed above.

Case 1: $n$ divides $998$

The three terms in the expression must be $(a, a, a)$ , as mentioned above, so the sum is $3a$ , which is divisible by $3$ . Therefore, the first case does not work.

Case 2: $n$ divides $999$

Because $n$ divides $999$ , the number of possibilities for $n$ is the same as the number of factors of $999$ . $999$ = $3^3 \cdot 37^1$ So, the total number of factors of $999$ is $4 \cdot 2 = 8$ .

However, we have to subtract $1$ , because the case $n = 1$ doesn't work, as mentioned previously. $8 - 1 = 7$

We now do the same for the third case.

Case 3: $n$ divides $1000$

Because $n$ divides $1000$ , the number of possibilities for $n$ is the same as the number of factors of $1000$ . $1000$ = $5^3 \cdot 2^3$ So, the total number of factors of $1000$ is $4 \cdot 4 = 16$ .

Again, we have to subtract $1$ , for the reason stated in Case 2. $16 - 1 = 15$

Case 4: $n$ divides none of { $998, 999, 1000$ }

If $n$ does not divide { $998, 999, 1000$ }, then the value of the terms of the above expression are $(a, a, a)$ . See if you can figure out why, although the explanation is similar to that of Case 1.

Therefore, the value of the expression is $3a$ , which is divisible by 3, so this case does not work.

Now that we have counted all of the cases, we add them.

$0 + 7 + 15 + 0 = 22$ , so the answer is $\boxed{\textbf{(A)}22}$ .

~dragonchomper

Video Solution

https://youtu.be/Ozp3k2464u4

~IceMatrix

Revision as of 22:10, 1 February 2020 (view source) Dragonchomper (talk \| contribs) (→‎Solution (Casework)) ← Older edit		Revision as of 22:11, 1 February 2020 (view source) Dragonchomper (talk \| contribs) (→‎Solution (Casework)) Newer edit →
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−	Now, we test the ~~three~~ cases listed above.	+	Now, we test the four cases listed above.

2020 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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Difference between revisions of "2020 AMC 10A Problems/Problem 22"

Revision as of 22:11, 1 February 2020

Contents

Problem

Solution (Casework)

Video Solution

See Also