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If we start off with one cell which goes on to divide 84 times, how many cells will there be after the final division?? Surely after it divides once we have 2 cells and since one division had already taken place, it’s 2^83?
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#2
I am pretty sure it is 2^84. We can simplify the problem by considering only dividing three times. After one division you have two cells, after two, 4 cells, and after three, 8 cells - which is 2^3. It follows that if you divide 84 times, you produce 2^84 cells.
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(Original post by anosmianAcrimony)
I am pretty sure it is 2^84. We can simplify the problem by considering only dividing three times. After one division you have two cells, after two, 4 cells, and after three, 8 cells - which is 2^3. It follows that if you divide 84 times, you produce 2^84 cells.
I am pretty sure it is 2^84. We can simplify the problem by considering only dividing three times. After one division you have two cells, after two, 4 cells, and after three, 8 cells - which is 2^3. It follows that if you divide 84 times, you produce 2^84 cells.
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#4
(Original post by Wolfram Alpha)
If we start off with one cell which goes on to divide 84 times, how many cells will there be after the final division?? Surely after it divides once we have 2 cells and since one division had already taken place, it’s 2^83?
If we start off with one cell which goes on to divide 84 times, how many cells will there be after the final division?? Surely after it divides once we have 2 cells and since one division had already taken place, it’s 2^83?
for example
if a cell divides 5 times
2^5 +32
after the first division there is 2 cells
after the second division there is 4 cells
after the third division there is 8 cells
after the fourth division there is 16 cells
after the fiftch division there is 32 cells
do you see what i mean?
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#5
(Original post by Wolfram Alpha)
Unfortunately this makes complete sense to me which means I’ve lost those two marks in my biology exam
Unfortunately this makes complete sense to me which means I’ve lost those two marks in my biology exam
On the bright side, 2^84 is somewhere around 10^25, and using the average weight of a cell, that would be in the order of thousands of tons of biomass - so the question was completely unrealistic.
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(Original post by 13dsaivaines)
No because the first division needs to be included in the 84.
for example
if a cell divides 5 times
2^5 +32
after the first division there is 2 cells
after the second division there is 4 cells
after the third division there is 8 cells
after the fourth division there is 16 cells
after the fiftch division there is 32 cells
do you see what i mean?
No because the first division needs to be included in the 84.
for example
if a cell divides 5 times
2^5 +32
after the first division there is 2 cells
after the second division there is 4 cells
after the third division there is 8 cells
after the fourth division there is 16 cells
after the fiftch division there is 32 cells
do you see what i mean?
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#8
(Original post by Wolfram Alpha)
Does this mean 2^83 is correct?
Does this mean 2^83 is correct?
because 5^4 would = 16 and that is not the answer to 5 divisions
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#10
This sounds like such a nice question. Shame my biology paper included stuff that wasn’t included in the specification 
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#11
(Original post by Wolfram Alpha)
If we start off with one cell which goes on to divide 84 times, how many cells will there be after the final division?? Surely after it divides once we have 2 cells and since one division had already taken place, it’s 2^83?
If we start off with one cell which goes on to divide 84 times, how many cells will there be after the final division?? Surely after it divides once we have 2 cells and since one division had already taken place, it’s 2^83?
2, 4, 8, ...
Then the number of cells after the nth division will be
2^n
So the answer is 2^84
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(Original post by //Destiel)
This sounds like such a nice question. Shame my biology paper included stuff that wasn’t included in the specification
This sounds like such a nice question. Shame my biology paper included stuff that wasn’t included in the specification
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(Original post by Notnek)
There are 2 cells after the first division, then 4 after the second etc. so your sequence is
2, 4, 8, ...
Then the number of cells after the nth division will be
2^n
So the answer is 2^84
There are 2 cells after the first division, then 4 after the second etc. so your sequence is
2, 4, 8, ...
Then the number of cells after the nth division will be
2^n
So the answer is 2^84
becausxe surely when we have 2 cells, 1 division has already occurred so we only have 83 divisions left? Can't believe I'm an A Levels maths student 
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#14
(Original post by Wolfram Alpha)
If we start off with one cell which goes on to divide 84 times, how many cells will there be after the final division?? Surely after it divides once we have 2 cells and since one division had already taken place, it’s 2^83?
If we start off with one cell which goes on to divide 84 times, how many cells will there be after the final division?? Surely after it divides once we have 2 cells and since one division had already taken place, it’s 2^83?
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#15
(Original post by Wolfram Alpha)
I've aceepted it's 2^84 but to b honest I still don't get why
becausxe surely when we have 2 cells, 1 division has already occurred so we only have 83 divisions left? Can't believe I'm an A Levels maths student 
I've aceepted it's 2^84 but to b honest I still don't get why
becausxe surely when we have 2 cells, 1 division has already occurred so we only have 83 divisions left? Can't believe I'm an A Levels maths student
.
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