How many three-digit numbers are divisible By 7
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1st 3 digit no. divisible by 7=105
Last 3 digit no. divisible by 7=994
Therefore AP: 105,112,119,.......,994
a=105 d=7 an=994 n=?
an=a+(n-1)d
994=105+(n-1)7
994-105=7n-7
896=7n
n=128
So,no.of 3 digit numbers divisible by 7 is 128
Hope it helps:)
Last 3 digit no. divisible by 7=994
Therefore AP: 105,112,119,.......,994
a=105 d=7 an=994 n=?
an=a+(n-1)d
994=105+(n-1)7
994-105=7n-7
896=7n
n=128
So,no.of 3 digit numbers divisible by 7 is 128
Hope it helps:)
anu554:
thank u so much
Answered by
0
Answer:
1st 3 digit no. divisible by 7=105
Last 3 digit no. divisible by 7=994
Therefore AP: 105,112,119,.......,994
a=105 d=7 an=994 n=?
an=a+(n-1)d
994=105+(n-1)7
994-105=7n-7
896=7n
n=128
So,no.of 3 digit numbers divisible by 7 is 128
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