How many 3 digits positive integers exist that when divided by 7 leave a remainder of 0

128 numbers are divisible by 7 that leave remainder 0

128

Step-by-step explanation:

Let's find the first 3 digit number divisible by 7.

7(12)=84

7(13)=91

7(14)=98

7(15)=105

So our first 3 digit number divisible by 7 is 105.

Let's figure out the last 3 digit number divisible by 7.

7(200)=1400

Skip a lot

7(150)=1050

Skip a little more

7(140)=980

Skipped too much

7(145)=1015

7(144)=1008

7(143)=1001

7(142)=994

So there are 142-15+1 multiples of 7 that are 3 digits long.

142-15+1

142-14

128

i would go with the first one

step-by-step explanation:

hope this

b

step-by-step explanation: