how many 3 digit number devisible by 5
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Answered by
1
first 3 digit no divisible by 5 is 100 so a=100
last no divisible by 5 is 905 so tnn=905
d=5
tn=a+(n-1)d
995=100+(n-1)5
995=100+5n-5
995=95+5n
995-95=5n
900=5n
900/5=n
n=180
last no divisible by 5 is 905 so tnn=905
d=5
tn=a+(n-1)d
995=100+(n-1)5
995=100+5n-5
995=95+5n
995-95=5n
900=5n
900/5=n
n=180
priteshrathwa:
sorry my mistake
what's the answer
o ya
now ok
Answered by
3
Hii there ....heres the answee...
A.P=100,105,110,………..,995
Therefore,a=100
d=5
a+(n-1)d=995
100+(n-1)(5)=995
(n-1)(5)=895
(n-1)=179
n=180
Therefore ,there are 180 three digit numbers which are divisible by 5
Mark the brain if i am ri8
A.P=100,105,110,………..,995
Therefore,a=100
d=5
a+(n-1)d=995
100+(n-1)(5)=995
(n-1)(5)=895
(n-1)=179
n=180
Therefore ,there are 180 three digit numbers which are divisible by 5
Mark the brain if i am ri8
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