Brain Teasers
Quocorrian Math 3
Bound by your promise to stay on Quocorria until you've mastered all of their numbering systems, you find yourself still stuck on Quocorria.
Now, while attending a Quocorrian mathematical seminar, the speaker mentions yet another form of math. After the speech, you approach him with questions on this system. Being like the rest of the Quocorrians you've met, he only gives you a sheet of paper, this time with six problems on it, and leaves.
Here's the paper:
2+2=11
6-4=28
4*5=100
10/5=10
3-4=13
4+3*5-6=9497
So, knowing that those are all true, what's 3*6?
Now, while attending a Quocorrian mathematical seminar, the speaker mentions yet another form of math. After the speech, you approach him with questions on this system. Being like the rest of the Quocorrians you've met, he only gives you a sheet of paper, this time with six problems on it, and leaves.
Here's the paper:
2+2=11
6-4=28
4*5=100
10/5=10
3-4=13
4+3*5-6=9497
So, knowing that those are all true, what's 3*6?
Hint
This system is not commutative. 3+2 is not the same as 2+3.The Order of Operations is the same as our own.
1+1=5.5
Answer
90.In this system, you multiply the rightmost number by 1, the second rightmost number by 10, the third by 100, and so on. Solve the problem like that, and divide by two.
So 2+2 would look like (20+2)/2.
4+3*5-6 would look like (4000+300*50-6)/2
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Comments
that's a really good one! awesome
Heh, heh...
Didn't get it...
Didn't get it...
Mar 09, 2005
Sorry but I don't think the solution is proper. Try this isntead:
for addition/subtraction
multiply the first number times 5 then divide the second number by 2
example
2+2=11 > (2*5)+(2/2)=11
for multiplication
multiply the first number times 5
example
4x5=100 (4x5)x5=100
This works for everything but the last equation in the problem, including the answer.
Cheers,
Matt
for addition/subtraction
multiply the first number times 5 then divide the second number by 2
example
2+2=11 > (2*5)+(2/2)=11
for multiplication
multiply the first number times 5
example
4x5=100 (4x5)x5=100
This works for everything but the last equation in the problem, including the answer.
Cheers,
Matt
While these are acceptable, they are conditional, depending on the operation.
Also, the reason why your addition/subtraction method works is because it's the exact same thing as my original. The original can be represented by:
(10a+b)/2=c where a is the first number, b is the second, and c is the answer in a two number problem. Multiply both sides by two, and you get 10a+b=2c. Divide by two, and you get 5a+(b/2)=c. (You could of just factored, but that thought came to me too late )
Also, the reason why your addition/subtraction method works is because it's the exact same thing as my original. The original can be represented by:
(10a+b)/2=c where a is the first number, b is the second, and c is the answer in a two number problem. Multiply both sides by two, and you get 10a+b=2c. Divide by two, and you get 5a+(b/2)=c. (You could of just factored, but that thought came to me too late )
Mar 10, 2005
Ah! Missed the divide by 2 part at the end of the solution. Thanks for clearing it up.
No problem. When reading your response, and trying to figure out why yours worked, I almost missed it too
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