Question 1205868
<br>
Tutor @ikleyn did not solve the problem using trial and error.  She correctly showed that, since the number leaves remainder 2 when divided by either 4 or 7, it also leaves a remainder 2 when divided by 4*7 = 28.  Therefore the number is of the form 28k+2 for some integer k.<br>
From there, since we are looking for the smallest number of that form which also leaves a remainder of 5 when divided by 9, the solution is easily found simply by starting with the smallest integer values of k and trying them until a number is found which satisfies all of the conditions.  That is not trial and error; it is a logical solution method.<br>
If a more formal mathematical finish to the problem is needed for some reason, then some easy arithmetic mod 9 can be used.<br>
For k=1, the number is 28+2 = 30; mod(30,9) = 3.<br>
Each time we increase k by 1, the number increases by 28; mod(28,9) = 1.  So each time we increase k by 1, mod(N,9) increases by 1.<br>
For the number we are looking for, we want mod(N,9) = 5.<br>
For k=1, we had mod(N,9) = 3; and for each increase of 1 in k, mod(N,9) increases by 1.  To get mod(N,9) = 5, we need to increase 3 by 1 two times, so the answer is when k = 1+2 = 3.<br>
And that gives us (formally) the answer N = 28(3)+2 = 86<br>