how many 3 digit numbers are divisible by 6 which always leave remainder 1?
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a=103
d=6
an=997
an=a+(n-1)d
997=103+(n-1)6
997-103=6n-6
894=6n-6
894+6=6n
900=6n
n=900/6
n=150
hence,there are 150 3-digit no. that are divisible by 6 and leave remainder 1....
d=6
an=997
an=a+(n-1)d
997=103+(n-1)6
997-103=6n-6
894=6n-6
894+6=6n
900=6n
n=900/6
n=150
hence,there are 150 3-digit no. that are divisible by 6 and leave remainder 1....
amimariam1:
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