Easy math puzzles
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51 members have voted
November 16th, 2025 at 3:06:19 AM
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How can you enclose all nine circles with two additional squares?
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
November 16th, 2025 at 8:45:15 AM
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Quote: Wizard
Move one coin to create two lines of four coins each.
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It depends on how you define "line."
I assume the answer is, put the blue coin on top of the red one, but then the four coins in each row are, strictly speaking, not in a "line."
I assume the answer is, put the blue coin on top of the red one, but then the four coins in each row are, strictly speaking, not in a "line."
November 16th, 2025 at 9:14:00 AM
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Quote: Wizard
Draw three paths, connecting A to A, B to B, and C to C. The paths may not cross nor go outside the box.
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I've always seen this done differently. The three squares on the bottom represent electricity, water and gas. You need to hook up all utilities by connecting each house to all three, without crossing any lines.
The older I get, the better I recall things that never happened
November 16th, 2025 at 9:24:52 AM
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Quote: billryanI've always seen this done differently. The three squares on the bottom represent electricity, water and gas. You need to hook up all utilities by connecting each house to all three, without crossing any lines.
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That's a different problem, with the added condition that no line could go through another house or utility
It was proven centuries ago that this is impossible, regardless of where the three houses and three utilities are located.
November 16th, 2025 at 9:46:52 AM
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Quote: ThatDonGuyQuote: billryanI've always seen this done differently. The three squares on the bottom represent electricity, water and gas. You need to hook up all utilities by connecting each house to all three, without crossing any lines.
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That's a different problem, with the added condition that no line could go through another house or utilityIt was proven centuries ago that this is impossible, regardless of where the three houses and three utilities are located.
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A sage Pack leader once offered $100( in 1967 dollars) reward for the solution to a bus full of Cub Scouts at the start of a three-hour bus ride. It's a great babysitting tool.
I actually saw a twist on this where they substituted telephone for gas, which makes it possible.
The older I get, the better I recall things that never happened
November 17th, 2025 at 4:43:05 PM
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Quote: ThatDonGuy
It depends on how you define "line."
I assume the answer is, put the blue coin on top of the red one, but then the four coins in each row are, strictly speaking, not in a "line."
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I agree!
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
November 17th, 2025 at 4:45:21 PM
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Draw two squares to enclose all nine circles. You may count the existing square as well.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
November 18th, 2025 at 8:39:52 AM
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A square is divided into 16 boxes.
Each box is numbered 1-16.
Place the numbers so that each column, horizontal, vertical, and diagonal, adds up to 34.
This is considered the most powerful of the magic squares, and solving it will protect you from the plague.
For a smaller blessing, divide the square into 9 parts and place the numbers so each adds up to 15.
Jewish people revere this square as it represents the name of God.
Each box is numbered 1-16.
Place the numbers so that each column, horizontal, vertical, and diagonal, adds up to 34.
This is considered the most powerful of the magic squares, and solving it will protect you from the plague.
For a smaller blessing, divide the square into 9 parts and place the numbers so each adds up to 15.
Jewish people revere this square as it represents the name of God.
The older I get, the better I recall things that never happened
November 18th, 2025 at 6:38:12 PM
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Quote: billryanA square is divided into 16 boxes.
Each box is numbered 1-16.
Place the numbers so that each column, horizontal, vertical, and diagonal, adds up to 34.
This is considered the most powerful of the magic squares, and solving it will protect you from the plague.
For a smaller blessing, divide the square into 9 parts and place the numbers so each adds up to 15.
Jewish people revere this square as it represents the name of God.
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How about a mega-blessing?
Here is how you can create a magic square of any odd size:
Put the number 1 in the center column of the top square.
For each number, the next number's square is the one diagonally to the upper right.
If you are already at the top row, put it on the bottom row of the next column to the right.
If you are already at the right column, put it on the left column of the next row up.
If you are in the upper right corner, or the square up and to the right is occupied, put the next number directly below the current one.
There was once an official Guinness World Record for the largest magic square, and I think it was "only" 125 x 125.
November 18th, 2025 at 6:46:24 PM
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Quote: Wizard
Draw two squares to enclose all nine circles. You may count the existing square as well.
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I am not entirely sure it is possible "as drawn."
I assume the intended answer is, the vertices of the first drawn square are on the large square - either the midpoints of each side, or moved slightly clockwise around the square - then the vertices of the second drawn square are the midpoints of the first drawn square.
However, given the shape of the square and the size and location of the nine circles, I am not entirely sure it can be done here. Every time I try, I intersect one of the circles.
November 18th, 2025 at 6:59:17 PM
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Quote: Wizard
How can you enclose all nine circles with two additional squares?
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Being the circles look they already are enclosed by a square, I made an assumption that it means drawing two squares that isolate each circle from the others.
November 19th, 2025 at 3:58:56 AM
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Quote: AutomaticMonkey
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I agree. My artwork was not so great either as evidently the lines as they should have been crossed some of the circles.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
November 19th, 2025 at 10:27:11 AM
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not a math puzzle but a rebus the nursing home passed out but we can't figure it out.anypne have any ideas to the solution?
November 19th, 2025 at 10:57:49 AM
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I had to look the phrase up, as I had never heard it before:
"A wise head on young shoulders"
(hay, wise, head, on (switch), young (someone's third birthday), shoulders)
"A wise head on young shoulders"
(hay, wise, head, on (switch), young (someone's third birthday), shoulders)
November 19th, 2025 at 6:56:19 PM
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Find the area of the green region.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
November 19th, 2025 at 8:53:08 PM
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Area=
37
November 19th, 2025 at 9:51:42 PM
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Quote: acesideArea=
37
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I agree. However, for full credit, you must show your work.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
November 20th, 2025 at 6:22:18 AM
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⬜🟩⬜🟨⬜ TRIAL
⬜⬜⬜🟨⬜ CONES
🟩🟩🟩⬜🟩 GRAPE
🟩🟩🟩🟩🟩 GRAVE
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
November 20th, 2025 at 7:10:43 AM
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Quote: WizardQuote: acesideArea=
37
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I agree. However, for full credit, you must show your work.
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Let a be the horizontal offset of the intersection from the midpoint;
Let b be the vertical offset of the intersection from the midpoint;
Let c be a half side length of the square.
The area of the two bottom quadrants is Eq.1:
24+31+c b=2c c;
The area of the left two quadrants is Eq.2:
24+30+c a=2c c;
The area of the left bottom quadrant is Eq.3:
24+c b/2+c a/2=c c.
Solving Eq.1-Eq.2, we have Eq.4:
c=1/(a-b);
Solving Eq.1+Eq.2, we have Eq.5:
109+c (a+b)=4 c c.
Therefore, plugging Eq.4 into Eq. 5, we have Eq.6:
109(a-b)(a-b)+(a+b)(a-b)=4.
Plugging Eq.4 into Eq.3, we have Eq.7:
24(a-b)(a-b)+(1/2)(a+b)(a-b)=1.
Solving this equation set of Eq.6 and Eq.7, we easily have Eq.8
(a-b)(a-b)=2/61;
And Eq.9:
(a+b)(a-b)=26/61.
Using Eq.8 and Eq.9, we have the unknow area, calculated as
(a+c)(b+c)-a(b+c)/2-b(a+c)/2=[(a+b)(a-b)/2+1]/[(a-b)(a-b)]=[(26/61)/2+1]/(2/61)=37.
Let b be the vertical offset of the intersection from the midpoint;
Let c be a half side length of the square.
The area of the two bottom quadrants is Eq.1:
24+31+c b=2c c;
The area of the left two quadrants is Eq.2:
24+30+c a=2c c;
The area of the left bottom quadrant is Eq.3:
24+c b/2+c a/2=c c.
Solving Eq.1-Eq.2, we have Eq.4:
c=1/(a-b);
Solving Eq.1+Eq.2, we have Eq.5:
109+c (a+b)=4 c c.
Therefore, plugging Eq.4 into Eq. 5, we have Eq.6:
109(a-b)(a-b)+(a+b)(a-b)=4.
Plugging Eq.4 into Eq.3, we have Eq.7:
24(a-b)(a-b)+(1/2)(a+b)(a-b)=1.
Solving this equation set of Eq.6 and Eq.7, we easily have Eq.8
(a-b)(a-b)=2/61;
And Eq.9:
(a+b)(a-b)=26/61.
Using Eq.8 and Eq.9, we have the unknow area, calculated as
(a+c)(b+c)-a(b+c)/2-b(a+c)/2=[(a+b)(a-b)/2+1]/[(a-b)(a-b)]=[(26/61)/2+1]/(2/61)=37.
Last edited by: aceside on Nov 20, 2025
November 20th, 2025 at 7:23:07 AM
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Quote: WizardI agree. However, for full credit, you must show your work.
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Let 2s be the side length of the square
The blue area = ab + 1/2 b (s - a) + 1/2 a (s - b)
= 1/2 as + 1/2 bs
The green area = (2s - a)(2s - b) - 1/2 (s - a)(2s - b) - 1/2 (s - b)(2s - a)
= 4 s2 - 2as - 2bs + ab - s2 + 1/2 bs + as - 1/2 ab - s2 + 1/2 as + bs - 1/2 ab
= 2 s2 - 1/2 as - 1/2 bs
The sum of the blue and green areas = 2 s2; since the sum of all four areas = the area of the square = 4 s2, the sum of the red and yellow areas = 2 s2 = the sum of the blue and green areas
The green area = 30 + 31 - 24 = 37
November 20th, 2025 at 11:28:14 AM
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You're installing something on a wall and need to locate studs behind the drywall. You don’t have a studfinder so your only option is to start drilling holes until you find one.
What's the most efficient way to find the first stud?
Assumptions: Studs are spaced at sixteen inches center to center and are 1.5 inches wide. The drill bit is very thin so ignore it's diameter for calculation purposes
What's the most efficient way to find the first stud?
Assumptions: Studs are spaced at sixteen inches center to center and are 1.5 inches wide. The drill bit is very thin so ignore it's diameter for calculation purposes
It’s all about making that GTA
November 20th, 2025 at 1:51:47 PM
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Quote: Ace2You're installing something on a wall and need to locate studs behind the drywall. You don’t have a studfinder so your only option is to start drilling holes until you find one.
What's the most efficient way to find the first stud?
Assumptions: Studs are spaced at sixteen inches center to center and are 1.5 inches wide. The drill bit is very thin so ignore it's diameter for calculation purposes
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I assume "efficient" means having the lowest expected number of holes drilled.
Drill a spot chosen at random, then make each subsequent spot 1 1/2 inches to the right of the previous spot. Any move of more than 1 1/2 inches runs the risk of skipping over a stud. The gap between studs is 14 1/2 inches, so if your first 10 holes are misses, there is no need to make an 11th as there must be a stud 1 1/2 inches to the right of the 11th hole.
Actually, the distance between holes should be as slightly less than 1 1/2 inches as you can, to prevent the possibility of drilling a hole on both sides of a stud.
