Asked by Linda
Snap to It...
I snap my fingers now, in one second, I smap them again. I wait two seconds and snap them a third time. I wait 4 seconds before snapping them again. Then I wait eight, then sixteen, and the pattern continues. The interval between each snap doubles each time. How many times will I snap my fingers during the next year.
I snap my fingers now, in one second, I smap them again. I wait two seconds and snap them a third time. I wait 4 seconds before snapping them again. Then I wait eight, then sixteen, and the pattern continues. The interval between each snap doubles each time. How many times will I snap my fingers during the next year.
Answers
Answered by
Reiny
so you are really asking ...
1+2+4+8+16 + .... = number of seconds in a year
left side is the sum of a GS with
a = 1, r = 2
seconds in a year = 365(24)(60)960) = 31536000
sum(n) of GS = a(r^n - 1)/(r-1)
= a(2^r -1)/1
2^r - 1 = 31536000
2^r = 31536001
r log2 = log 31536001
r = 24.9
by the time you snap 25 times, the year would be over, so you can only snap 24 times
check:
24 snaps --- > 24 terms of 1+2+4+ ..
= 1(2^24 - 1)/(2-1)2 = 16777215 < less than seconds in a year
25 snaps ---> 25 terms of 1+2+4+...
= 2^25 -1 = 33554431 > seconds in a year
1+2+4+8+16 + .... = number of seconds in a year
left side is the sum of a GS with
a = 1, r = 2
seconds in a year = 365(24)(60)960) = 31536000
sum(n) of GS = a(r^n - 1)/(r-1)
= a(2^r -1)/1
2^r - 1 = 31536000
2^r = 31536001
r log2 = log 31536001
r = 24.9
by the time you snap 25 times, the year would be over, so you can only snap 24 times
check:
24 snaps --- > 24 terms of 1+2+4+ ..
= 1(2^24 - 1)/(2-1)2 = 16777215 < less than seconds in a year
25 snaps ---> 25 terms of 1+2+4+...
= 2^25 -1 = 33554431 > seconds in a year
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