Number theory question with floor function

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Define $[a]$ as the largest integer not greater than $a$. For example, $left[frac113right]=3$. Given the function
$$f(x)=left[frac x7right]left[frac37xright],$$
where $x$ is an integer such that $1le xle45$, how many values can $f(x)$ assume?

A. $1$ B. $3$ C. $4$ D. $5$ E. $6$

I have attempted this question by brute force however I am looking for a cleaner, more systematic approach.

contest-math floor-function

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edited Jul 23 at 3:21

Math Lover

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asked Jul 23 at 2:45

Isaac Greene

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up vote
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down vote

favorite

1

Define $[a]$ as the largest integer not greater than $a$. For example, $left[frac113right]=3$. Given the function
$$f(x)=left[frac x7right]left[frac37xright],$$
where $x$ is an integer such that $1le xle45$, how many values can $f(x)$ assume?

A. $1$ B. $3$ C. $4$ D. $5$ E. $6$

I have attempted this question by brute force however I am looking for a cleaner, more systematic approach.

contest-math floor-function

share|cite|improve this question

edited Jul 23 at 3:21

Math Lover

12.3k21232

asked Jul 23 at 2:45

Isaac Greene

15911

add a comment | 

up vote
7
down vote

favorite

1

up vote
7
down vote

favorite

1

1

Define $[a]$ as the largest integer not greater than $a$. For example, $left[frac113right]=3$. Given the function
$$f(x)=left[frac x7right]left[frac37xright],$$
where $x$ is an integer such that $1le xle45$, how many values can $f(x)$ assume?

A. $1$ B. $3$ C. $4$ D. $5$ E. $6$

I have attempted this question by brute force however I am looking for a cleaner, more systematic approach.

contest-math floor-function

share|cite|improve this question

edited Jul 23 at 3:21

Math Lover

12.3k21232

asked Jul 23 at 2:45

Isaac Greene

15911

Define $[a]$ as the largest integer not greater than $a$. For example, $left[frac113right]=3$. Given the function
$$f(x)=left[frac x7right]left[frac37xright],$$
where $x$ is an integer such that $1le xle45$, how many values can $f(x)$ assume?

A. $1$ B. $3$ C. $4$ D. $5$ E. $6$

I have attempted this question by brute force however I am looking for a cleaner, more systematic approach.

contest-math floor-function

share|cite|improve this question

edited Jul 23 at 3:21

Math Lover

12.3k21232

asked Jul 23 at 2:45

Isaac Greene

15911

share|cite|improve this question

share|cite|improve this question

edited Jul 23 at 3:21

Math Lover

12.3k21232

edited Jul 23 at 3:21

Math Lover

12.3k21232

edited Jul 23 at 3:21

Math Lover

12.3k21232

12.3k21232

asked Jul 23 at 2:45

Isaac Greene

15911

asked Jul 23 at 2:45

Isaac Greene

15911

asked Jul 23 at 2:45

Isaac Greene

15911

15911

add a comment | 

add a comment | 

2 Answers
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First, $f(x) le left[fracx7cdotfrac37xright] = 5$, and $f(x) ge 0$. Therefore, $f(x)$ can assume a maximum of $6$ values: $0,1,2,3,4,5$.

Second, $f(x)=1$ implies $left[fracx7right]=1$ and $left[frac37xright]=1$. The first condition implies $7 le x le 13$, while the latter requires $19 le x le 37$, which is impossible. So $f(x) neq 1$.

Checking first few values of $x$ reveals that $f(x)$ can assume the values $2,3,4,5$. Specifically, for $1 le x le 6$, we have $f(x)=0$ as $[x/7]=0$. For $x=7$, $f(x) = 5$; for $x=8,9$, $f(x)=4$; for $x=10, 11, 12$, $f(x)=3$, and for $x=13$, $f(x)=2$.

Consequently, the number of values $f(x)$ can assume is $5$.

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answered Jul 23 at 3:18

Math Lover

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We start from $x=1$ and work up. At this point, the left floor function returns zero so $f(x)=0$.

• This continues until $x=7$, whereupon the left floor is 1 and the right floor is 5, so $f(x)=5$.


• $lfloor37/8rfloor$ and $lfloor37/9rfloor$ are 4; $lfloor37/10rfloor$ to $lfloor37/12rfloor$ are 3; $lfloor37/13$ is 2. In all these cases the left floor expression is 1, so $f(x)=4,3,2$.


• At $x=14$ the left floor becomes 2. The right floor remains 2 ($f(x)=4$) until $x=19$ when it becomes 1 ($f(x)=2$).


• At $x=21$ the left floor becomes 3 ($f(x)=3$). The right floor remains 1, continuing through $x=28$ ($f(x)=4$) and $x=35$ ($f(x)=5$) to $x=37$, when the right floor reaches 0 and $f(x)=0$ for the remaining values of $x$ (to 45).

We find that $f(x)$ may assume the values $0,2,3,4,5$ – five distinct values. The answer is D.

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answered Jul 23 at 3:04

Parcly Taxel

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
7
down vote



accepted

First, $f(x) le left[fracx7cdotfrac37xright] = 5$, and $f(x) ge 0$. Therefore, $f(x)$ can assume a maximum of $6$ values: $0,1,2,3,4,5$.

Second, $f(x)=1$ implies $left[fracx7right]=1$ and $left[frac37xright]=1$. The first condition implies $7 le x le 13$, while the latter requires $19 le x le 37$, which is impossible. So $f(x) neq 1$.

Checking first few values of $x$ reveals that $f(x)$ can assume the values $2,3,4,5$. Specifically, for $1 le x le 6$, we have $f(x)=0$ as $[x/7]=0$. For $x=7$, $f(x) = 5$; for $x=8,9$, $f(x)=4$; for $x=10, 11, 12$, $f(x)=3$, and for $x=13$, $f(x)=2$.

Consequently, the number of values $f(x)$ can assume is $5$.

share|cite|improve this answer

answered Jul 23 at 3:18

Math Lover

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up vote
7
down vote



accepted

First, $f(x) le left[fracx7cdotfrac37xright] = 5$, and $f(x) ge 0$. Therefore, $f(x)$ can assume a maximum of $6$ values: $0,1,2,3,4,5$.

Second, $f(x)=1$ implies $left[fracx7right]=1$ and $left[frac37xright]=1$. The first condition implies $7 le x le 13$, while the latter requires $19 le x le 37$, which is impossible. So $f(x) neq 1$.

Checking first few values of $x$ reveals that $f(x)$ can assume the values $2,3,4,5$. Specifically, for $1 le x le 6$, we have $f(x)=0$ as $[x/7]=0$. For $x=7$, $f(x) = 5$; for $x=8,9$, $f(x)=4$; for $x=10, 11, 12$, $f(x)=3$, and for $x=13$, $f(x)=2$.

Consequently, the number of values $f(x)$ can assume is $5$.

share|cite|improve this answer

answered Jul 23 at 3:18

Math Lover

12.3k21232

add a comment | 

up vote
7
down vote



accepted

up vote
7
down vote



accepted

First, $f(x) le left[fracx7cdotfrac37xright] = 5$, and $f(x) ge 0$. Therefore, $f(x)$ can assume a maximum of $6$ values: $0,1,2,3,4,5$.

Second, $f(x)=1$ implies $left[fracx7right]=1$ and $left[frac37xright]=1$. The first condition implies $7 le x le 13$, while the latter requires $19 le x le 37$, which is impossible. So $f(x) neq 1$.

Checking first few values of $x$ reveals that $f(x)$ can assume the values $2,3,4,5$. Specifically, for $1 le x le 6$, we have $f(x)=0$ as $[x/7]=0$. For $x=7$, $f(x) = 5$; for $x=8,9$, $f(x)=4$; for $x=10, 11, 12$, $f(x)=3$, and for $x=13$, $f(x)=2$.

Consequently, the number of values $f(x)$ can assume is $5$.

share|cite|improve this answer

answered Jul 23 at 3:18

Math Lover

12.3k21232

First, $f(x) le left[fracx7cdotfrac37xright] = 5$, and $f(x) ge 0$. Therefore, $f(x)$ can assume a maximum of $6$ values: $0,1,2,3,4,5$.

Second, $f(x)=1$ implies $left[fracx7right]=1$ and $left[frac37xright]=1$. The first condition implies $7 le x le 13$, while the latter requires $19 le x le 37$, which is impossible. So $f(x) neq 1$.

Checking first few values of $x$ reveals that $f(x)$ can assume the values $2,3,4,5$. Specifically, for $1 le x le 6$, we have $f(x)=0$ as $[x/7]=0$. For $x=7$, $f(x) = 5$; for $x=8,9$, $f(x)=4$; for $x=10, 11, 12$, $f(x)=3$, and for $x=13$, $f(x)=2$.

Consequently, the number of values $f(x)$ can assume is $5$.

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answered Jul 23 at 3:18

Math Lover

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answered Jul 23 at 3:18

Math Lover

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up vote
3
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We start from $x=1$ and work up. At this point, the left floor function returns zero so $f(x)=0$.

• This continues until $x=7$, whereupon the left floor is 1 and the right floor is 5, so $f(x)=5$.


• $lfloor37/8rfloor$ and $lfloor37/9rfloor$ are 4; $lfloor37/10rfloor$ to $lfloor37/12rfloor$ are 3; $lfloor37/13$ is 2. In all these cases the left floor expression is 1, so $f(x)=4,3,2$.


• At $x=14$ the left floor becomes 2. The right floor remains 2 ($f(x)=4$) until $x=19$ when it becomes 1 ($f(x)=2$).


• At $x=21$ the left floor becomes 3 ($f(x)=3$). The right floor remains 1, continuing through $x=28$ ($f(x)=4$) and $x=35$ ($f(x)=5$) to $x=37$, when the right floor reaches 0 and $f(x)=0$ for the remaining values of $x$ (to 45).

We find that $f(x)$ may assume the values $0,2,3,4,5$ – five distinct values. The answer is D.

share|cite|improve this answer

answered Jul 23 at 3:04

Parcly Taxel

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up vote
3
down vote

We start from $x=1$ and work up. At this point, the left floor function returns zero so $f(x)=0$.

• This continues until $x=7$, whereupon the left floor is 1 and the right floor is 5, so $f(x)=5$.


• $lfloor37/8rfloor$ and $lfloor37/9rfloor$ are 4; $lfloor37/10rfloor$ to $lfloor37/12rfloor$ are 3; $lfloor37/13$ is 2. In all these cases the left floor expression is 1, so $f(x)=4,3,2$.


• At $x=14$ the left floor becomes 2. The right floor remains 2 ($f(x)=4$) until $x=19$ when it becomes 1 ($f(x)=2$).


• At $x=21$ the left floor becomes 3 ($f(x)=3$). The right floor remains 1, continuing through $x=28$ ($f(x)=4$) and $x=35$ ($f(x)=5$) to $x=37$, when the right floor reaches 0 and $f(x)=0$ for the remaining values of $x$ (to 45).

We find that $f(x)$ may assume the values $0,2,3,4,5$ – five distinct values. The answer is D.

share|cite|improve this answer

answered Jul 23 at 3:04

Parcly Taxel

33.5k136588

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up vote
3
down vote

up vote
3
down vote

We start from $x=1$ and work up. At this point, the left floor function returns zero so $f(x)=0$.

• This continues until $x=7$, whereupon the left floor is 1 and the right floor is 5, so $f(x)=5$.


• $lfloor37/8rfloor$ and $lfloor37/9rfloor$ are 4; $lfloor37/10rfloor$ to $lfloor37/12rfloor$ are 3; $lfloor37/13$ is 2. In all these cases the left floor expression is 1, so $f(x)=4,3,2$.


• At $x=14$ the left floor becomes 2. The right floor remains 2 ($f(x)=4$) until $x=19$ when it becomes 1 ($f(x)=2$).


• At $x=21$ the left floor becomes 3 ($f(x)=3$). The right floor remains 1, continuing through $x=28$ ($f(x)=4$) and $x=35$ ($f(x)=5$) to $x=37$, when the right floor reaches 0 and $f(x)=0$ for the remaining values of $x$ (to 45).

We find that $f(x)$ may assume the values $0,2,3,4,5$ – five distinct values. The answer is D.

share|cite|improve this answer

answered Jul 23 at 3:04

Parcly Taxel

33.5k136588

We start from $x=1$ and work up. At this point, the left floor function returns zero so $f(x)=0$.

• This continues until $x=7$, whereupon the left floor is 1 and the right floor is 5, so $f(x)=5$.


• $lfloor37/8rfloor$ and $lfloor37/9rfloor$ are 4; $lfloor37/10rfloor$ to $lfloor37/12rfloor$ are 3; $lfloor37/13$ is 2. In all these cases the left floor expression is 1, so $f(x)=4,3,2$.


• At $x=14$ the left floor becomes 2. The right floor remains 2 ($f(x)=4$) until $x=19$ when it becomes 1 ($f(x)=2$).


• At $x=21$ the left floor becomes 3 ($f(x)=3$). The right floor remains 1, continuing through $x=28$ ($f(x)=4$) and $x=35$ ($f(x)=5$) to $x=37$, when the right floor reaches 0 and $f(x)=0$ for the remaining values of $x$ (to 45).

We find that $f(x)$ may assume the values $0,2,3,4,5$ – five distinct values. The answer is D.

share|cite|improve this answer

answered Jul 23 at 3:04

Parcly Taxel

33.5k136588

share|cite|improve this answer

share|cite|improve this answer

answered Jul 23 at 3:04

Parcly Taxel

33.5k136588

answered Jul 23 at 3:04

Parcly Taxel

33.5k136588

answered Jul 23 at 3:04

Parcly Taxel

33.5k136588

33.5k136588

add a comment | 

add a comment | 

 

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