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Old 04-26-2011, 05:07 PM   #16
Hyperephania
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Default Re: Is this math problem solvable?

Strictly you could also "reverse" the 6 and get a 9:
49+35+12=96
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Old 04-26-2011, 05:28 PM   #17
Crystallas
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Default Re: Is this math problem solvable?

It's possible. I wont ruin the fun for you guys.
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Old 04-26-2011, 05:37 PM   #18
Myth
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Default Re: Is this math problem solvable?

Quote:
Originally Posted by Bano114
It might be a trick question?

14+56+23+3?

I combined the numbers to make another number and only used the number I made once. I dont know if combining counts as "using" the number or if it means you can only use it once in the addition?

I'm probably wrong but its just a thought.

That makes sense. If it said you can only use each digit once, then it would be more clear, but as it is written, it seems as if it could be interpreted your way.
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Old 04-26-2011, 08:31 PM   #19
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Default Re: Is this math problem solvable?

Quote:
Originally Posted by Crystallas
It's possible. I wont ruin the fun for you guys.

I'll call your bluff
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Old 04-26-2011, 08:38 PM   #20
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Default Re: Is this math problem solvable?

Quote:
Originally Posted by Crystallas
It's possible. I wont ruin the fun for you guys.

I can't see how based on the "rules" we have to follow. I'll give it one more go ahead later, but like others said, the closest I got so far is 93.

You're talking to the former Captain of his Math Team back in HS......oh yeah....... ........i gots all the ladies.........jealous?
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Old 04-26-2011, 09:42 PM   #21
arkain
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Default Re: Is this math problem solvable?

Quote:
Originally Posted by Hawker
53+42+1=96

Pwned

****...just saw AND 6

What if you shove the 6 as an exponent to the 1? I mean you are still only really adding.

53+42+1^6 = 96

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Old 04-26-2011, 09:51 PM   #22
Bano114
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Default Re: Is this math problem solvable?

Quote:
Originally Posted by arkain
What if you shove the 6 as an exponent to the 1? I mean you are still only really adding.

53+42+1^6 = 96


1 raised to the 6 is multiplying 1 by its self 6 times. Multiplying.
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Old 04-26-2011, 09:55 PM   #23
arkain
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Default Re: Is this math problem solvable?

Quote:
Originally Posted by Bano114
1 raised to the 6 is multiplying 1 by its self 6 times. Multiplying.

yeah i know, but it's implicit, not explicit
its a trick question, thats the best i could come up with that could somehow be justified as a trick lol
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Old 04-26-2011, 10:02 PM   #24
gotbacon23
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Default Re: Is this math problem solvable?

Quote:
Originally Posted by GatorKid117
If were looked at it from a trick question standpoint, it says to use 1-6 and only use them once but it doesn't specifically say your only limited to those numbers. So hypothetically, I guess you could use 7? I may be looking too much into this, I don't know.

i think this is the key.

Quote:
Originally Posted by Sicknote

Use the numbers 1,2,3,4,5, AND 6 to get 96 using only addition, and only using each number once. You can combine the numbers to make another number (for example 1 and 2 to make 12). Remember you can only use the number once.


it says you have to use 1,2, 3, 4, 5, 6 once each but it doesn't say ONLY use 1, 2, 3, 4, 5, 6. it actually puts no limit on how many times you can use the digits 0, 7, 8, 9 so i would assume there could be many answers.

12+45+3+6 = 66, 66 + 8 + 8 + 7 + 7 = 96

21+34 + 6 + 5 = 66, 66 + 8 + 8 + 7 + 7 = 96

36 + 24 + 1 + 5 = 66, 66 + 8 + 8 + 7 +7 =96
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Old 04-26-2011, 10:24 PM   #25
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Default Re: Is this math problem solvable?

Quote:
Originally Posted by arkain

53+42+1^6 = 96


I think this is the answer. Well done.
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