Posted: 7/5/2016 11:18:00 AM EDT
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256, 16, 4, 2, 6561, 81, 9, 3, ___, 256
What is the missing number, I cant figure it out |
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256, 16, 4, 2, 6561, 81, 9, 3, , 256
is 2^8, 2^4, 2^2, 2^1, 3^8, 3^4, 3^2, 3^1, ???, 2^8 Given the series as is with no information if it's the start, middle, end, or complete, 2^16 is your best bet. ETA: The beginning could be: 2^1, 3^1, 2^2, 2^1, 3^2, 3^1, 2^4, 2^2, 2^1, 3^4, 3^2, 3^1, given, and then continues by adding a doubled exponent at the beginning of each "2" and "3" smaller series. |
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Quoted:
Think of it like this (256, 16, 4, 2,) 2 (6561, 81, 9, 3,) 3 ( ___, 256, 16, 4) 4 etc. so 256^2 = 65536 is the logical entry The next set is (390625, 625, 25, 5) 5 Why not my example? Yours would start "(1, 1, 1, 1) 1", presumably. But we don't have that information. You've also included ", 16, 4" in the last line of your series, but that's not included in the original series. |
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Quoted:
Why not my example? Yours would start "(1, 1, 1, 1) 1", presumably. But we don't have that information. You've also included ", 16, 4" in the last line of your series, but that's not included in the original series. Quoted:
Quoted:
Think of it like this (256, 16, 4, 2,) 2 (6561, 81, 9, 3,) 3 ( ___, 256, 16, 4) 4 etc. so 256^2 = 65536 is the logical entry The next set is (390625, 625, 25, 5) 5 Why not my example? Yours would start "(1, 1, 1, 1) 1", presumably. But we don't have that information. You've also included ", 16, 4" in the last line of your series, but that's not included in the original series. OK You win I'll be on my way now... |
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Quoted:
OK You win I'll be on my way now... Quoted:
Quoted:
Quoted:
Think of it like this (256, 16, 4, 2,) 2 (6561, 81, 9, 3,) 3 ( ___, 256, 16, 4) 4 etc. so 256^2 = 65536 is the logical entry The next set is (390625, 625, 25, 5) 5 Why not my example? Yours would start "(1, 1, 1, 1) 1", presumably. But we don't have that information. You've also included ", 16, 4" in the last line of your series, but that's not included in the original series. You win I'll be on my way now... 
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Quoted:
Noooooo....... it was an honest question. I assumed you read mine and dismissed it due to some error/inconsistency. I mean, I can't disagree with yours either, I just means that with the limited information, we both arrived at the same answer using different paths. Sorry, I thought you were hurt I posted an alternative because you were "in first". 
I just saw it differently and posted an alternative way of thinking about it. You're correct the first set would be all 1s, and I simply added the other terms to set number 4 to illustrate how the stacking of each subset should appear logically. However, in your example I think the final 2^8 and therefore 2^16 works only because of the coincidence of 4 being 2^2. Because the next set is not 3^16, it's 5^8. your final terms should be 4^4 and therefore 4^8. |
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Quoted:
Sorry, I thought you were hurt I posted an alternative because you were "in first".
I just saw it differently and posted an alternative way of thinking about it. You're correct the first set would be all 1s, and I simply added the other terms to set number 4 to illustrate how the stacking of each subset should appear logically. However, in your example I think the final 2^8 and therefore 2^16 works only because of the coincidence of 4 being 2^2. Because the next set is not 3^16, it's 5^8. your final terms should be 4^4 and therefore 4^8. Quoted:
Quoted:
Noooooo....... it was an honest question. I assumed you read mine and dismissed it due to some error/inconsistency. I mean, I can't disagree with yours either, I just means that with the limited information, we both arrived at the same answer using different paths. Sorry, I thought you were hurt I posted an alternative because you were "in first".
I just saw it differently and posted an alternative way of thinking about it. You're correct the first set would be all 1s, and I simply added the other terms to set number 4 to illustrate how the stacking of each subset should appear logically. However, in your example I think the final 2^8 and therefore 2^16 works only because of the coincidence of 4 being 2^2. Because the next set is not 3^16, it's 5^8. your final terms should be 4^4 and therefore 4^8. I totally see that. I just went with 2, 3, repeating, instead of 2, 3, 4, 5, continuing. If there were one preceding number or three succeeding numbers, we could know for sure which way the series was going. Of course, it's still 65,536 either way. 
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