The easiest way to solve this is to write out for the larger divisors those numbers up to 100 where the dividing conditions are satisfied. For 5 it's 9, 14, 24, 29,...99 and for 4 it's 7, 11, 15, 19,...,99. So for 4 and 5 we only have n, the number we're dividing, = 19, 39, 59, 79 and 99. The only number on this list that divides by 3 with remainder 2 is 59, which, divided by 7 has a remainder of 3.
Another way of solving it is to take the numbers 4 and 5 and multiply them together=20. Subtract 1 from this and we get 19. 4 has remainder 3 and 5 has remainder 4 (both requirements of the question) when divided into 19 and this happens for every number separated by 20. Similarly, if we multiply 3, 4 and 5 together we get 60. Subtract 1 to get 59 and we meet all the requirements. This would be repeated every 60 numbers but we only need to go up to 100.