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How is my math:

Subduction Zone

Veteran Member
I happened to get the same answer:

PEMDAS is what I used as well 196 - 182*0.5 = 196 - 91 = 105. But let's not stop there:
105 = 7*5*3 = 7!!


I saw a short video on this where a math guy copied the same thing from someone else and when he said the username, and I forgot the whole name, but I figured my initial reaction was wrong too because the user name had "factorial" as part of it. I knew there was something funny going on. It was the first that I had heard of double factorial myself. I just wanted to share it in the same way since my reaction was the same as all of yours.
 

Debater Slayer

Vipassana
Staff member
Premium Member
196 - 182*0.5 = 7!!

Did I make any errors?

Yes.

The order of operations is PEMDAS (or BODMAS, if you use British English). This means the result is 196 - 91 = 105.

Also, 7! Is 5040, if you were going for that too, so it still wouldn't be the answer. :D

Edit:

Okay, I see from a later reply of yours that you went for a double factorial, which I didn't catch. Good one!
 

The Hammer

Skald
Premium Member
What do the exclamation points represent? Anything other than excitement in your answer?

A (!) Symbol means to do a factorial on the previous number in a math problem..

 

exchemist

Veteran Member
196 - 182*0.5 = 7!!

Did I make any errors?
I remember learning BODMAS at school in about 1964. My son learnt a slightly different mnemonic for the same thing in about 2014. So yeah 105.

Unless you are going to tell us that the * symbol for some reason is to be applied in a different order from x, in which I look forward to your explanation.
 

sun rise

The world is on fire
Premium Member
Google Gemini choked utterly. ChatGPT tried and failed

Q: In what numeric base is this equation true 196 - 182*0.5 = 7

A: To find out in what numeric base this equation is true, let's express it generically in base bb:
196b−182b×0.5b=7b196b−182b×0.5b=7b
First, let's convert the decimal numbers to base bb:
19610=1×b2+9×b1+6×b019610=1×b2+9×b1+6×b0 18210=1×b2+8×b1+2×b018210=1×b2+8×b1+2×b0 0.510=5×b−10.510=5×b−1
Now, let's substitute these values into the equation and solve for bb:
(1×b2+9×b1+6×b0)−(1×b2+8×b1+2×b0)×(5×b−1)=7b(1×b2+9×b1+6×b0)−(1×b2+8×b1+2×b0)×(5×b−1)=7b
Simplify the equation:
(1×b2+9×b1+6×b0)−(1×b2+8×b1+2×b0)×(5×b−1)=7b(1×b2+9×b1+6×b0)−(1×b2+8×b1+2×b0)×(5×b−1)=7b (1×b2+9×b1+6×b0)−(5×b−1×(1×b2+8×b1+2×b0))=7b(1×b2+9×b1+6×b0)−(5×b−1×(1×b2+8×b1+2×b0))=7b (1×b2+9×b1+6×b0)−(5×b−1×b2+40×b0+10×b−1)=7b(1×b2+9×b1+6×b0)−(5×b−1×b2+40×b0+10×b−1)=7b 1×b2+9×b1+6×b0−5×b1−40×b0−10×b−1=7b1×b2+9×b1+6×b0−5×b1−40×b0−10×b−1=7b
Now, equate the coefficients of corresponding powers of bb on both sides of the equation:
1b2−5b1+6b0−10b−1−40b0=71b2−5b1+6b0−10b−1−40b0=7
b2−5b+6−10b−40=7b2−5b+6−b10−40=7
b2−5b−34−10b=0b2−5b−34−b10=0
This equation can be solved to find the value(s) of bb that make the original equation true. However, without further context or constraints, it's difficult to determine the exact base in which the equation holds true. It's worth noting that solutions for bb may not always be unique, and some bases may not be practical or meaningful in certain contexts.
 

Subduction Zone

Veteran Member
I remember learning BODMAS at school in about 1964. My son learnt a slightly different mnemonic for the same thing in about 2014. So yeah 105.

Unless you are going to tell us that the * symbol for some reason is to be applied in a different order from x, in which I look forward to your explanation.
Nah, it is the two, not just one, exclamation points.
 

Subduction Zone

Veteran Member
Google Gemini choked utterly. ChatGPT tried and failed

Q: In what numeric base is this equation true 196 - 182*0.5 = 7

A: To find out in what numeric base this equation is true, let's express it generically in base bb:
196b−182b×0.5b=7b196b−182b×0.5b=7b
First, let's convert the decimal numbers to base bb:
19610=1×b2+9×b1+6×b019610=1×b2+9×b1+6×b0 18210=1×b2+8×b1+2×b018210=1×b2+8×b1+2×b0 0.510=5×b−10.510=5×b−1
Now, let's substitute these values into the equation and solve for bb:
(1×b2+9×b1+6×b0)−(1×b2+8×b1+2×b0)×(5×b−1)=7b(1×b2+9×b1+6×b0)−(1×b2+8×b1+2×b0)×(5×b−1)=7b
Simplify the equation:
(1×b2+9×b1+6×b0)−(1×b2+8×b1+2×b0)×(5×b−1)=7b(1×b2+9×b1+6×b0)−(1×b2+8×b1+2×b0)×(5×b−1)=7b (1×b2+9×b1+6×b0)−(5×b−1×(1×b2+8×b1+2×b0))=7b(1×b2+9×b1+6×b0)−(5×b−1×(1×b2+8×b1+2×b0))=7b (1×b2+9×b1+6×b0)−(5×b−1×b2+40×b0+10×b−1)=7b(1×b2+9×b1+6×b0)−(5×b−1×b2+40×b0+10×b−1)=7b 1×b2+9×b1+6×b0−5×b1−40×b0−10×b−1=7b1×b2+9×b1+6×b0−5×b1−40×b0−10×b−1=7b
Now, equate the coefficients of corresponding powers of bb on both sides of the equation:
1b2−5b1+6b0−10b−1−40b0=71b2−5b1+6b0−10b−1−40b0=7
b2−5b+6−10b−40=7b2−5b+6−b10−40=7
b2−5b−34−10b=0b2−5b−34−b10=0
This equation can be solved to find the value(s) of bb that make the original equation true. However, without further context or constraints, it's difficult to determine the exact base in which the equation holds true. It's worth noting that solutions for bb may not always be unique, and some bases may not be practical or meaningful in certain contexts.
One exclamation point: factorial. Two exclamation points: double factorial.
 
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