Will two kangaroos ever meet after making same number of jumps?
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There are two kangaroos on an x-axis ready to jump in the positive direction (i.e, toward positive infinity). The first kangaroo starts at location $x_1$ and moves at a rate of $v_1$ meters per jump. The second kangaroo starts at location $x_2$ and moves at a rate of $v_2$ meters per jump. Given the starting locations and movement rates for each kangaroo, can you determine if they'll ever land at the same location at the same time?
Input Format
A single line of four space-separated integers denoting the respective values of $x_1$, $v_1$, $x_2$, and $v_2$.
Constraints
- $0 le x_1 < x_2$
- $1 le v_1$
- $1 le v_2$
Output Format
Print YES if they can land on the same location at the same time; otherwise, print NO.
Note: The two kangaroos must land at the same location after making the same number of jumps.
Sample Input 0
0 3 4 2
Sample Output 0
YES
Explanation 0
The two kangaroos jump through the following sequence of locations:
- 0 3 6 9 12
- 4 6 8 10 12
Thus, the kangaroos meet after 4 jumps and we print YES.
Sample Input 1
0 2 5 3
Sample Output 1
NO
Explanation 1
The second kangaroo has a starting location that is ahead (further to the right) of the first kangaroo's starting location (i.e., $x_2 > x_1$). Because the second kangaroo moves at a faster rate (meaning $v_2 > v_1$) and is already ahead of the first kangaroo, the first kangaroo will never be able to catch up. Thus, we print NO.
Note: I searched for the answer and got this puzzle here but without answer :( so had to ask.
mathematics
edited 18 hours ago
Glorfindel
12.7k34879
asked 19 hours ago
Govind Prajapati
1655
add a comment |
up vote
1
down vote
favorite
There are two kangaroos on an x-axis ready to jump in the positive direction (i.e, toward positive infinity). The first kangaroo starts at location $x_1$ and moves at a rate of $v_1$ meters per jump. The second kangaroo starts at location $x_2$ and moves at a rate of $v_2$ meters per jump. Given the starting locations and movement rates for each kangaroo, can you determine if they'll ever land at the same location at the same time?
Input Format
A single line of four space-separated integers denoting the respective values of $x_1$, $v_1$, $x_2$, and $v_2$.
Constraints
- $0 le x_1 < x_2$
- $1 le v_1$
- $1 le v_2$
Output Format
Print YES if they can land on the same location at the same time; otherwise, print NO.
Note: The two kangaroos must land at the same location after making the same number of jumps.
Sample Input 0
0 3 4 2
Sample Output 0
YES
Explanation 0
The two kangaroos jump through the following sequence of locations:
- 0 3 6 9 12
- 4 6 8 10 12
Thus, the kangaroos meet after 4 jumps and we print YES.
Sample Input 1
0 2 5 3
Sample Output 1
NO
Explanation 1
The second kangaroo has a starting location that is ahead (further to the right) of the first kangaroo's starting location (i.e., $x_2 > x_1$). Because the second kangaroo moves at a faster rate (meaning $v_2 > v_1$) and is already ahead of the first kangaroo, the first kangaroo will never be able to catch up. Thus, we print NO.
Note: I searched for the answer and got this puzzle here but without answer :( so had to ask.
mathematics
edited 18 hours ago
Glorfindel
12.7k34879
asked 19 hours ago
Govind Prajapati
1655
What is the source of this puzzle?
– Dr Xorile
13 hours ago
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
There are two kangaroos on an x-axis ready to jump in the positive direction (i.e, toward positive infinity). The first kangaroo starts at location $x_1$ and moves at a rate of $v_1$ meters per jump. The second kangaroo starts at location $x_2$ and moves at a rate of $v_2$ meters per jump. Given the starting locations and movement rates for each kangaroo, can you determine if they'll ever land at the same location at the same time?
Input Format
A single line of four space-separated integers denoting the respective values of $x_1$, $v_1$, $x_2$, and $v_2$.
Constraints
- $0 le x_1 < x_2$
- $1 le v_1$
- $1 le v_2$
Output Format
Print YES if they can land on the same location at the same time; otherwise, print NO.
Note: The two kangaroos must land at the same location after making the same number of jumps.
Sample Input 0
0 3 4 2
Sample Output 0
YES
Explanation 0
The two kangaroos jump through the following sequence of locations:
- 0 3 6 9 12
- 4 6 8 10 12
Thus, the kangaroos meet after 4 jumps and we print YES.
Sample Input 1
0 2 5 3
Sample Output 1
NO
Explanation 1
The second kangaroo has a starting location that is ahead (further to the right) of the first kangaroo's starting location (i.e., $x_2 > x_1$). Because the second kangaroo moves at a faster rate (meaning $v_2 > v_1$) and is already ahead of the first kangaroo, the first kangaroo will never be able to catch up. Thus, we print NO.
Note: I searched for the answer and got this puzzle here but without answer :( so had to ask.
mathematics
edited 18 hours ago
Glorfindel
12.7k34879
asked 19 hours ago
Govind Prajapati
1655
There are two kangaroos on an x-axis ready to jump in the positive direction (i.e, toward positive infinity). The first kangaroo starts at location $x_1$ and moves at a rate of $v_1$ meters per jump. The second kangaroo starts at location $x_2$ and moves at a rate of $v_2$ meters per jump. Given the starting locations and movement rates for each kangaroo, can you determine if they'll ever land at the same location at the same time?
Input Format
A single line of four space-separated integers denoting the respective values of $x_1$, $v_1$, $x_2$, and $v_2$.
Constraints
- $0 le x_1 < x_2$
- $1 le v_1$
- $1 le v_2$
Output Format
Print YES if they can land on the same location at the same time; otherwise, print NO.
Note: The two kangaroos must land at the same location after making the same number of jumps.
Sample Input 0
0 3 4 2
Sample Output 0
YES
Explanation 0
The two kangaroos jump through the following sequence of locations:
- 0 3 6 9 12
- 4 6 8 10 12
Thus, the kangaroos meet after 4 jumps and we print YES.
Sample Input 1
0 2 5 3
Sample Output 1
NO
Explanation 1
The second kangaroo has a starting location that is ahead (further to the right) of the first kangaroo's starting location (i.e., $x_2 > x_1$). Because the second kangaroo moves at a faster rate (meaning $v_2 > v_1$) and is already ahead of the first kangaroo, the first kangaroo will never be able to catch up. Thus, we print NO.
Note: I searched for the answer and got this puzzle here but without answer :( so had to ask.
mathematics
mathematics
edited 18 hours ago
Glorfindel
12.7k34879
asked 19 hours ago
Govind Prajapati
1655
edited 18 hours ago
Glorfindel
12.7k34879
asked 19 hours ago
Govind Prajapati
1655
edited 18 hours ago
Glorfindel
12.7k34879
edited 18 hours ago
Glorfindel
12.7k34879
edited 18 hours ago
Glorfindel
12.7k34879
12.7k34879
asked 19 hours ago
Govind Prajapati
1655
asked 19 hours ago
Govind Prajapati
1655
asked 19 hours ago
Govind Prajapati
1655
1655
What is the source of this puzzle?
– Dr Xorile
13 hours ago
add a comment |
What is the source of this puzzle?
– Dr Xorile
13 hours ago
What is the source of this puzzle?
– Dr Xorile
13 hours ago
What is the source of this puzzle?
– Dr Xorile
13 hours ago
add a comment |
1 Answer
1
active
oldest
votes
up vote
6
down vote
accepted
They'll meet if and only if
$v_1 > v_2$ (so that kangaroo 1 catches up)
and
$v_1 - v_2 | x_2 - x_1$, here | means 'is a divisor of'.
Why?
After $n in Bbb{N}$ jumps, kangaroo 1 will be at position $x_1 + n v_1$ and kangaroo 2 at $x_2 + n v_2$. Now, if
$$x_1 + n v_1 = x_2 + n v_2$$
$$n v_1 - n v_2 = x_2 - x_1$$
$$n (v_1 - v_2) = x_2 - x_1$$
$$n = frac{x_2 - x_1}{v_1 - v_2}$$
This fraction is an integer if and only if $v_1 - v_2$ divides $x_2 - x_1$.
answered 18 hours ago
Glorfindel
12.7k34879
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
accepted
They'll meet if and only if
$v_1 > v_2$ (so that kangaroo 1 catches up)
and
$v_1 - v_2 | x_2 - x_1$, here | means 'is a divisor of'.
Why?
After $n in Bbb{N}$ jumps, kangaroo 1 will be at position $x_1 + n v_1$ and kangaroo 2 at $x_2 + n v_2$. Now, if
$$x_1 + n v_1 = x_2 + n v_2$$
$$n v_1 - n v_2 = x_2 - x_1$$
$$n (v_1 - v_2) = x_2 - x_1$$
$$n = frac{x_2 - x_1}{v_1 - v_2}$$
This fraction is an integer if and only if $v_1 - v_2$ divides $x_2 - x_1$.
answered 18 hours ago
Glorfindel
12.7k34879
add a comment |
up vote
6
down vote
accepted
They'll meet if and only if
$v_1 > v_2$ (so that kangaroo 1 catches up)
and
$v_1 - v_2 | x_2 - x_1$, here | means 'is a divisor of'.
Why?
After $n in Bbb{N}$ jumps, kangaroo 1 will be at position $x_1 + n v_1$ and kangaroo 2 at $x_2 + n v_2$. Now, if
$$x_1 + n v_1 = x_2 + n v_2$$
$$n v_1 - n v_2 = x_2 - x_1$$
$$n (v_1 - v_2) = x_2 - x_1$$
$$n = frac{x_2 - x_1}{v_1 - v_2}$$
This fraction is an integer if and only if $v_1 - v_2$ divides $x_2 - x_1$.
answered 18 hours ago
Glorfindel
12.7k34879
add a comment |
up vote
6
down vote
accepted
up vote
6
down vote
accepted
They'll meet if and only if
$v_1 > v_2$ (so that kangaroo 1 catches up)
and
$v_1 - v_2 | x_2 - x_1$, here | means 'is a divisor of'.
Why?
After $n in Bbb{N}$ jumps, kangaroo 1 will be at position $x_1 + n v_1$ and kangaroo 2 at $x_2 + n v_2$. Now, if
$$x_1 + n v_1 = x_2 + n v_2$$
$$n v_1 - n v_2 = x_2 - x_1$$
$$n (v_1 - v_2) = x_2 - x_1$$
$$n = frac{x_2 - x_1}{v_1 - v_2}$$
This fraction is an integer if and only if $v_1 - v_2$ divides $x_2 - x_1$.
answered 18 hours ago
Glorfindel
12.7k34879
They'll meet if and only if
$v_1 > v_2$ (so that kangaroo 1 catches up)
and
$v_1 - v_2 | x_2 - x_1$, here | means 'is a divisor of'.
Why?
After $n in Bbb{N}$ jumps, kangaroo 1 will be at position $x_1 + n v_1$ and kangaroo 2 at $x_2 + n v_2$. Now, if
$$x_1 + n v_1 = x_2 + n v_2$$
$$n v_1 - n v_2 = x_2 - x_1$$
$$n (v_1 - v_2) = x_2 - x_1$$
$$n = frac{x_2 - x_1}{v_1 - v_2}$$
This fraction is an integer if and only if $v_1 - v_2$ divides $x_2 - x_1$.
answered 18 hours ago
Glorfindel
12.7k34879
edited 18 hours ago
answered 18 hours ago
Glorfindel
12.7k34879
answered 18 hours ago
Glorfindel
12.7k34879
answered 18 hours ago
Glorfindel
12.7k34879
12.7k34879
add a comment |
add a comment |
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What is the source of this puzzle?
– Dr Xorile
13 hours ago