how many 4 digit number divisible by 8
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four digit numbers are 1000,1001,1002,1003,1004,1005.........9996,9997,9998,9999.
now 1000,1008,1016.....9992 are divisible by 8
so, applying arithmetic progression
number of term(n) =?
first term(a) = 1000
difference(d)=8
last term= 9992
therefore 9992=a+(n-1)d
9992=1000+(n-1)×8
(n-1)×8=8996
n-1=8992/8
n-1=1124
n=1124+1
n=1125
so there are 1125 four digit numbers divisible by 8
now 1000,1008,1016.....9992 are divisible by 8
so, applying arithmetic progression
number of term(n) =?
first term(a) = 1000
difference(d)=8
last term= 9992
therefore 9992=a+(n-1)d
9992=1000+(n-1)×8
(n-1)×8=8996
n-1=8992/8
n-1=1124
n=1124+1
n=1125
so there are 1125 four digit numbers divisible by 8
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