Answer:
128 numbers are divisible by 7 that leave remainder 0
Answer:
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
