A frog located at (x,y), with both x and y integers, makes successive

Question

A frog located at (x,y), with both x and y integers, makes successive jumps of length 5 and always lands on points with integer coordinates. Suppose that the frog starts at (0,0) and ends at (1,0). What is the smallest possible number of jumps the frog makes?

A. 2
B. 3
C. 4
D. 5
E. 6

A. 2
B. 3
C. 4
D. 5
E. 6

GMATinsight · Answer

There will three jumps needed as shown in figure

First Jump: (0,0) to (3,4) Pink one

Second Jump (3,4) to (6,0) Green one

Third Jump (6,0) to (1,0) Orange one

Answer: Option B

abhishek31 · Answer

hi can anyone explain the approach to solving this question?

GMATinsight · Answer

It's already explained above. Let me know what doesn't make sense here..

Aderonke01 · Answer

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Chei · Answer

Guess I didn't even understand the question. GMATinsight  eraborate on your answer how you applied 5 to get the new coordinate.

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GMATinsight · Answer

The question meant that a frog jumps with jumo length 5 units but it lands at a point where the co-ordinates are integers.

Since 5 units is diagonal length from origin to (3,4) or (4,3) so these are the only two choices that frog has to jump at.

But if it jumped at (4,3) then the final condition may not be met hence we choose that the point jumped at is ((3,4)

next it again jumps 5 units away and finally it should be 1 unit away after the third jump so x-coordinate must be 6 units away before the third jump

hence points jumped at are listed as mentioned in above mentioned solution

abhishek31 · Answer

Hi, I wanted to understand, that how did you shortlist the points in the random plane? Lets say had it not been 5 points but 6 points jump, so how should i approach this question mathematically?

Pankaj0901 · Answer

Valid question. I guess it is more like a hit and trial attempt. Knowing pythagoras triplets (3-4-5) and having a basic understanding of coordinate geometry would help to chalk out possible options quickly. When you see a number 5 and unit 1 from the origin, clicking on mind the triplets (4=5-1, and then 3=4-1) would be natural provided you are familiar with basic triplets. And, trying the 5 unit leaps horizontally and vertically would not take you anywhere close to (1,0) and hence trying oblique leaps becomes the next option.

Hi, I wanted to understand, that how did you shortlist the points in the random plane? Lets say had it not been 5 points but 6 points jump, so how should i approach this question mathematically?[/quote]

Hi, I wanted to understand, that how did you shortlist the points in the random plane? Lets say had it not been 5 points but 6 points jump, so how should i approach this question mathematically?

Nups1324 · Answer

Hi GMATinsight It would be really helpful if you could show us how did you shortlist the point (3,4) See if the frog needs to be at (1,0) in the end, then it has to be at point (6,0) or (-4,0). Then only can a jump of 5 units will land it on (1,0). But how did you find (3,4)? I saw your attachment and I used this. Let our desirable coordinates be (x,y) then Distance between (0,0) & (x,y) = Distance between (6,0) & (x,y) = 5 Right.? Then by solving we get x as 3. Then Distance between (6,0) & (3,y) = 5 We get y as 2. Tagging others just in case Bunuel chetan2u yashikaaggarwal IanStewart ScottTargetTestPrep VeritasKarishma fskilnik Thank you

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GMATinsight · Answer

Length of Jump =5 and coordinates are integers. If you list out coordinates of points which have integers coordinates and are at a distance of 5 then you would find the following points in one quadrant and one of 4 axes (5,0) ((3,4) (4,3) Since frog has to make smallest possible jumps and has to end journey at (1,0) so you need to first devise the path that frog will follow. The devised path is shown in picture of my solution. I hope that help!

ScottTargetTestPrep · Answer

Solution:
 
Notice that the distance between (0, 0) and (3, 4) is 5. You can either apply the distance formula (which is √ ), or you can observe that the horizontal distance between (0, 0) and (3, 4) is 3 and the vertical distance between the two points is 4. Thus, the distance between the two points is equal to the hypotenuse of a right triangle where the legs have lengths 3 and 4; i.e. 5.

To understand the reason why (3, 4) was selected as the first point, notice that the frog can jump only to points (±5, 0), (0, ±5), (±3, ±4) and (±4, ±3). These are the only points with integer coordinates that have a distance of 5 from the origin.

Notice that (0, 0) -> (3, 4) -> (6, 0) -> (1, 0) is not the only three-step solution to this problem. Here are some alternate solutions:

(0, 0) -> (-5, 0) -> (-2, 4) -> (1, 0)

(0, 0) -> (-5, 0) -> (-2, -4) -> (1, 0)

(0, 0) -> (3, -4) -> (6, 0) -> (1, 0)

etc.

Finally, the distance between (6, 0) and (3, 2) is not 5. The distance between these two points is √ ) = √ ) = √(9 + 4) = √13.

KarishmaB · Answer

The length of the jump will be 5 and it must land on integer co-ordinates. It means the co-ordinates will form pythagorean triplets. (3, 4) and (4, 3) will give hypotenuse of 5 for diagonal jumps. Or the frog can make straight vertical or horizontal jumps to (5, 0), (0, 5) etc. Since it starts at (0, 0) and ends at (1, 0), we can stick to the first quadrant.

Now, the final change in x co-ordinate is an increase of 1 and in y co-ordinate there is no change. In no combination of 2 steps is that possible because y co-ordinates need to cancel each other but y co-ordinates are all different.

Next, let's consider if 3 jumps are possible. So to cancel off the y co-ordinates, we will need to make the same jump in different directions (up and down) and hopefully, we will be able to adjust the x co-ordinate.
So if the frog jumps 5 units (upwards and right) to reach (3, 4) and then another 5 units downwards and right to reach (6, 0), it can now jump back left 5 units to reach (1, 0).

Then, minimum 3 jumps are needed.

Answer (B)

Oppenheimer1945 · Answer

Initial Point (0,0)
Second point =(5,0)
3rd point will lie on circle (5+cos z, sin z)
If 4th pt is (1,0), dist=5

25=(4+cos z)^2+(sin z)^2
=> cos z=1, sin z=0
(point is clearly an integer with distance of 5 units from (1,0)), hence 3 jumps are enough

finisher009 · Answer

The 2 min Solution

1. Decode the Jump: A jump of length 5 landing on integer coordinates means the change in x and y for each jump (let’s call them ‘a’ and ‘b’) must be integers that satisfy the distance formula: a^2 + b^2 = 5^2, or a^2 + b^2 = 25. This is a classic Pythagorean triple. The only integer pairs for (a,b) are combinations of (+/-3, +/-4) and (+/-5, 0).

2. The Parity Hack (Odd/Even Trick): Look at the possible changes for x and y: {0, 3, 4, 5}. In every valid jump vector, one component is odd (3 or 5) and the other is even (0 or 4).
 • Start Point: (0, 0) -> (Even, Even)
 • After 1 jump: (Even, Even) + (Odd, Even) = (Odd, Even). The frog’s coordinates are now one odd, one even.
 • After 2 jumps: (Odd, Even) + (Odd, Even) = (Even, Even). The frog is back to having two even coordinates.
 • Destination: (1, 0) -> (Odd, Even)

3. Eliminate and Conclude: The frog can only be at a location with (Odd, Even) coordinates after an odd number of jumps. This instantly eliminates answer choices A (2 jumps), C (4 jumps), and E (6 jumps). The smallest possible answer remaining is B (3).

(Self-Correction/Verification: Is 3 actually possible? Yes. Consider the jumps: (+3, +4), (+3, -4), and (-5, 0). The total change in x is 3 + 3 - 5 = 1. The total change in y is 4 - 4 + 0 = 0. This path works, confirming 3 is the minimum.)

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A frog located at (x,y), with both x and y integers, makes successive

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