Will two kangaroos ever meet after making same number of jumps?
$begingroup$
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 Dec 13 '18 at 9:24
Glorfindel
13.8k35084
asked Dec 13 '18 at 9:10
Govind PrajapatiGovind Prajapati
1655
$endgroup$
add a comment |
$begingroup$
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 Dec 13 '18 at 9:24
Glorfindel
13.8k35084
asked Dec 13 '18 at 9:10
Govind PrajapatiGovind Prajapati
1655
$endgroup$
$begingroup$
What is the source of this puzzle?
$endgroup$
– Dr Xorile
Dec 13 '18 at 15:09
add a comment |
$begingroup$
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 Dec 13 '18 at 9:24
Glorfindel
13.8k35084
asked Dec 13 '18 at 9:10
Govind PrajapatiGovind Prajapati
1655
$endgroup$
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 Dec 13 '18 at 9:24
Glorfindel
13.8k35084
asked Dec 13 '18 at 9:10
Govind PrajapatiGovind Prajapati
1655
edited Dec 13 '18 at 9:24
Glorfindel
13.8k35084
asked Dec 13 '18 at 9:10
Govind PrajapatiGovind Prajapati
1655
edited Dec 13 '18 at 9:24
Glorfindel
13.8k35084
edited Dec 13 '18 at 9:24
Glorfindel
13.8k35084
edited Dec 13 '18 at 9:24
Glorfindel
13.8k35084
13.8k35084
asked Dec 13 '18 at 9:10
Govind PrajapatiGovind Prajapati
1655
asked Dec 13 '18 at 9:10
Govind PrajapatiGovind Prajapati
1655
asked Dec 13 '18 at 9:10
Govind PrajapatiGovind Prajapati
1655
1655
$begingroup$
What is the source of this puzzle?
$endgroup$
– Dr Xorile
Dec 13 '18 at 15:09
add a comment |
$begingroup$
What is the source of this puzzle?
$endgroup$
– Dr Xorile
Dec 13 '18 at 15:09
$begingroup$
What is the source of this puzzle?
$endgroup$
– Dr Xorile
Dec 13 '18 at 15:09
$begingroup$
What is the source of this puzzle?
$endgroup$
– Dr Xorile
Dec 13 '18 at 15:09
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 Dec 13 '18 at 9:16
GlorfindelGlorfindel
13.8k35084
$endgroup$
add a comment |
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1 Answer
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active
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1 Answer
1
active
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$begingroup$
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 Dec 13 '18 at 9:16
GlorfindelGlorfindel
13.8k35084
$endgroup$
add a comment |
$begingroup$
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 Dec 13 '18 at 9:16
GlorfindelGlorfindel
13.8k35084
$endgroup$
add a comment |
$begingroup$
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 Dec 13 '18 at 9:16
GlorfindelGlorfindel
13.8k35084
$endgroup$
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 Dec 13 '18 at 9:16
GlorfindelGlorfindel
13.8k35084
edited Dec 13 '18 at 9:20
answered Dec 13 '18 at 9:16
GlorfindelGlorfindel
13.8k35084
answered Dec 13 '18 at 9:16
GlorfindelGlorfindel
13.8k35084
answered Dec 13 '18 at 9:16
GlorfindelGlorfindel
13.8k35084
13.8k35084
add a comment |
add a comment |
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$begingroup$
What is the source of this puzzle?
$endgroup$
– Dr Xorile
Dec 13 '18 at 15:09