Question

43) how
many three-digit numbers
are divisible by 7?​

Answer 1

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

Answer 2

Step-by-step explanation:

first three digit = 105

last three digit = 994

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