Only Respond if you are OTAs: 101478, 103846, 104591, 104455

please check for an error. If you can improve the solution in its
clarity/ do so. Please try and use more words to explain the proofs/
make them more elegant if possible. YOU CAN STILL PICK UP THE CREDIT,
JUST AS LONG AS YOU CHECKED THE SOLUTION.



If there is something wrong with the solution, PLEASE REWRITE THE ENTIRE
SOLUTION.









(1) N=5k+2, because 2 candies left over if they try to divide N evenly
among 5 of them

(2) N-26 = 4m+1, because 1 candies left over if they try to divide N-26
evenly among 4 of them

(3) N-26-24 = 3h, because they can divide N-26-24 evenly into 3 parts



Therefore N=2+5k (1)

N=27+4m (2) and

N=50+3h, (3) where k, m and h are nonnegative integers

From (3), N >=50, so N=50,53,56,59,62,65,68,…

From (2) and the fact N>=50, N=51,55,59,63,67…

From (1) and the fact N>=50, N=52,57,62,67,…



So N=107 is the smallest value satisfying (1) (2) and (3)

The reason is because

N=50+3n_1 (1)

N=51+4n_2, (2)

N=52+5n_3, (3) where n_1,n_2,n_3 are integers

If they are equal, then

(2)-(1) ,we have 4*n_2+1=3*n_1 or 4*n_2=3*n_1 -1

(3)-(2), we have 5*n_3+1=4*n_2 , which tells us 4*n_2 = 5*n_3 + 1



So we need to find a smallest multiple of 4 which is a multiple of 3
minus 1, and also is a multiple of 5 plus 1. This smallest value is 56.

Therefore N=51+56=107, we have other values of N bigger than 107,

N=107+60k, k=0,1,2,…, where 60 is exactly the smallest common multiple
of 3,4 and 5.