Question 1141390
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He is adding 9 quarters every 3 days, or 3 quarters every day.<br>
A. At the end of day 2, he had added 2*3=6 quarters to the starting number.  Since he had 26 quarters at the end of day 2, he started with 26-6=20 quarters.<br>
B. 20, plus 3 more for each day: literally, y = 20+3x.  In slope-intercept form, y = 3x+20.<br>
C. The starting number of 20 is 2 more than a multiple of 3.  Since he adds 3 quarters each day, the total at the end of every day will be 2 more than a multiple of 3.  100 is 1 more than a multiple of 3, so he will never have 100 quarters at the end of any day.<br>
Note that instead of explaining part C in words, you could just show that the equation 3x+20=100 has a non-integer solution.