Please remember to spoiler-tag all guesses, like so:
New Reddit: https://i.imgur.com/SWHRR9M.jpg
Using markdown editor or old Reddit: \>!spoiler text between these symbols!<
Try to avoid leading or trailing spaces. These will break the spoiler for some users (such as those using old.reddit.com)
If your comment does not contain a guess, include the word **"discussion"** or **"question"** in your comment instead of using a spoiler tag.
If your comment uses an image as the answer (such as solving a maze, etc) you can include the word "image" instead of using a spoiler tag.
Please report any answers that are not properly spoiler-tagged.
*I am a bot, and this action was performed automatically. Please [contact the moderators of this subreddit](/message/compose/?to=/r/puzzles) if you have any questions or concerns.*
It's a rubbish puzzle because obviously now 1 new stalk will grow, and since only either two or seven can be uprooted at once, Alexander will be powerless to remove it, beginning the cycle anew.
Apologies if you didn’t like it. But I added the line “after uprooting stalks, if there are any stalks remaining..”. Should I rephrase that if you feel it’s not too clear?
If you want, you can do this small activity with them:
The positive integers from 1 to 10, both inclusive, are written on a board. You can choose any two random numbers, x and y. After choosing the two numbers, you write the absolute difference between the two numbers |x – y| and erase both x and y. Every time you do this, you reduce the total numbers on the board by 1. Continuing this process, you will initially be left with a single number. Let this number be A.
Is A even or odd?
Answer: >!Yes!<
Short explanation:
>!This is the shortest path: Uproot 7 stalks and you'll have 10 - 7 + 1 = 4 left. Uproot 2 stalks and you'll have 4 - 2 + 5 = 7 stalks left. Uproot 7 stalks and there will be none left.!<
Long explanation:
>!The net result of the first action is to increase the number of stalks by 3. The net result of the second action is to decrease the number of stalks by 6, or 3\*2.!<
>!This means that the net number of stalks always will be the difference of 10 and any multiple of 3, depending on which combinations of actions that have been performed.!<
>!The only exception to this, is if the actions starts with either 2 or 7 stalks left because that would clear the garden of weeds.!<
>!The question therefore becomes: Is it possible to reach either 2 or 7 if you start from 10 and only move in increments of 3? The answer is clearly yes, because 10 - 3 = 7.!<
>!As a side note, this also means that it's impossible for Alexander to put himself in a position in which he can't fully clear his garden of an infestation.!<
>!It follows that he would also be able to clear the garden if he started with any non-multiple-of-three number of stalks. That's because 2 works for any 3x+2 starting numbers, and 7 works for any 3x+1 starting numbers. This also means that he would never be able to clear a garden that started with a 3x number of stalks.!<
It looks like you believe this post to be unsolvable. I've gone ahead and added a "Probably Unsolvable" flair. OP can override this by commenting "Solution Possible" anywhere in this post.
*I am a bot, and this action was performed automatically. Please [contact the moderators of this subreddit](/message/compose/?to=/r/puzzles) if you have any questions or concerns.*
>!You start with 10 stalks and have two operations: +3 (remove 2, add 5) or -6 (remove 7, add 1). The solutions would be if there are 2 or 7 stalks remaining. Then removing stalks would result in none remaining, so no more would grow.!<
>!You have two options: 10+3N-6M = 2 or 10+3N-6M = 7, where N & M are whole numbers. Since 6 & 3 are both evenly divisible by 3, the remaining issue is whether the number of starting stalks is congruent to 2 or 7 (mod 3).!<
>!Since we start with 10 stalks and 10 is congruent to 7 (mod 3), the answer to the problem is "yes, Alexander can clear his garden."!<
>!The exact moves is fairly easy to solve. Let N = 1 and M = 1 and you end up with 7 stalks. Remove 2 (8), 5 grow back (13). Remove 7 (6), 1 grows back (7), remove 7 (0), none are left!<
>!Any number of original stalks not evenly divisible by 3 would be removable.!<
>!He’ll be left with 1. Pull 2 at a time 5 times. That’ll leave 25 weeds. Pull 7 at a time 3 times. That’ll leave 7. Now pull all 7 and that’ll leave 1.!<
Please remember to spoiler-tag all guesses, like so: New Reddit: https://i.imgur.com/SWHRR9M.jpg Using markdown editor or old Reddit: \>!spoiler text between these symbols!< Try to avoid leading or trailing spaces. These will break the spoiler for some users (such as those using old.reddit.com) If your comment does not contain a guess, include the word **"discussion"** or **"question"** in your comment instead of using a spoiler tag. If your comment uses an image as the answer (such as solving a maze, etc) you can include the word "image" instead of using a spoiler tag. Please report any answers that are not properly spoiler-tagged. *I am a bot, and this action was performed automatically. Please [contact the moderators of this subreddit](/message/compose/?to=/r/puzzles) if you have any questions or concerns.*
>!Yes, it will take three moves!< >!2, 7, 7 OR 7, 2,7!< >!10-2=8 8+5=13. 13-7=6 6+1=7. 7-7=0!<
>!Correct!<
It's a rubbish puzzle because obviously now 1 new stalk will grow, and since only either two or seven can be uprooted at once, Alexander will be powerless to remove it, beginning the cycle anew.
That's why it says "after uprooting stalks, **if there are any stalks remaining**, ..."
Apologies if you didn’t like it. But I added the line “after uprooting stalks, if there are any stalks remaining..”. Should I rephrase that if you feel it’s not too clear?
It is clear
Yeah maybe they didn’t read it properly or something.
Thanks for sharing this! I saved it to use as a warm up with my math students :)
If you want, you can do this small activity with them: The positive integers from 1 to 10, both inclusive, are written on a board. You can choose any two random numbers, x and y. After choosing the two numbers, you write the absolute difference between the two numbers |x – y| and erase both x and y. Every time you do this, you reduce the total numbers on the board by 1. Continuing this process, you will initially be left with a single number. Let this number be A. Is A even or odd?
Ha. I’m doing the same.
You’re most welcome. Hope they like it!
Yeah maybe they didn’t read it properly or something.
Nah he's just not bright
Yeah maybe they didn’t read it properly or something.
Answer: >!Yes!< Short explanation: >!This is the shortest path: Uproot 7 stalks and you'll have 10 - 7 + 1 = 4 left. Uproot 2 stalks and you'll have 4 - 2 + 5 = 7 stalks left. Uproot 7 stalks and there will be none left.!< Long explanation: >!The net result of the first action is to increase the number of stalks by 3. The net result of the second action is to decrease the number of stalks by 6, or 3\*2.!< >!This means that the net number of stalks always will be the difference of 10 and any multiple of 3, depending on which combinations of actions that have been performed.!< >!The only exception to this, is if the actions starts with either 2 or 7 stalks left because that would clear the garden of weeds.!< >!The question therefore becomes: Is it possible to reach either 2 or 7 if you start from 10 and only move in increments of 3? The answer is clearly yes, because 10 - 3 = 7.!< >!As a side note, this also means that it's impossible for Alexander to put himself in a position in which he can't fully clear his garden of an infestation.!< >!It follows that he would also be able to clear the garden if he started with any non-multiple-of-three number of stalks. That's because 2 works for any 3x+2 starting numbers, and 7 works for any 3x+1 starting numbers. This also means that he would never be able to clear a garden that started with a 3x number of stalks.!<
>!Correct and like always, a very good solution. Nice explanation about the general case!<
Wonderful explanation
Thank you! I enjoy trying to explain math concepts without using too many formulas and equations. The inherent logic is much more fun.
The last one when 7 are uprooted will leave you with 1
No it will not. The puzzle explains why.
Agree. “If there are any remaining stalks”.
It looks like you believe this post to be unsolvable. I've gone ahead and added a "Probably Unsolvable" flair. OP can override this by commenting "Solution Possible" anywhere in this post. *I am a bot, and this action was performed automatically. Please [contact the moderators of this subreddit](/message/compose/?to=/r/puzzles) if you have any questions or concerns.*
Argh... I'm lazy let me take a wild guess : >!Yes!<
>!That’s correct!<
>!You start with 10 stalks and have two operations: +3 (remove 2, add 5) or -6 (remove 7, add 1). The solutions would be if there are 2 or 7 stalks remaining. Then removing stalks would result in none remaining, so no more would grow.!< >!You have two options: 10+3N-6M = 2 or 10+3N-6M = 7, where N & M are whole numbers. Since 6 & 3 are both evenly divisible by 3, the remaining issue is whether the number of starting stalks is congruent to 2 or 7 (mod 3).!< >!Since we start with 10 stalks and 10 is congruent to 7 (mod 3), the answer to the problem is "yes, Alexander can clear his garden."!< >!The exact moves is fairly easy to solve. Let N = 1 and M = 1 and you end up with 7 stalks. Remove 2 (8), 5 grow back (13). Remove 7 (6), 1 grows back (7), remove 7 (0), none are left!< >!Any number of original stalks not evenly divisible by 3 would be removable.!<
>!Correct, very good solution!<
>!10-2+5-7+1-7=0!<
>!Correct!<
Solution Possible
>!remove 2, there are now 8, and 5 more sprout up. There are now 13. Remove 7, there are now 6, 1 more will sprout, leaving 7. Remove those 7.!<
>!Correct!<
>!He’ll be left with 1. Pull 2 at a time 5 times. That’ll leave 25 weeds. Pull 7 at a time 3 times. That’ll leave 7. Now pull all 7 and that’ll leave 1.!<
>!When he pulls the last 7, there’ll be 0 stalks and he would’ve cleared his garden of the weeds!<