N/2 = a + 1 or N = 2a + 2
N/3 = b + 1 or N = 3b + 3
N/4 = c + 1 or N = 4c + 4
N/5 = d or N = 5d
Can you take it from here?
Another example:
What is the least number that leaves a remainder of 3 when divided by 5, a remainder of 2 when divided by 4, a remainder of 1 when divided by 3, and a remainder of 0 when divided by 2?
From the problem statement, and letting the number we seek equal N, we can write
N/5 = A + 3/5 or N = 5A + 3
N/4 = B + 2/4 or N = 4B + 2
N/3 = C + 1/3 or N = 3C + 1
N/2 = D + 0 or N = 2D + 0
1--Combining the first two we get 4B - 5A = 1
2--Dividing through by 4 yields B - A - A/4 = 1/4 or (A + 1)/4 = B - A
3--(A + 1)/4 must equal an integer k making A = 4k - 1
4--Substituting back into (1) yields 4B - 20k + 5 = 1 or B = 5k - 1
5..k can be any integer from 1 on up
6--k.....1.....2.....3.....4.....5.....6
....A....3.....7....11....15...19...23
....B....4.....9....14....19...24...29
7--Combining the 2nd and 3rd expressions we get 3C - 4B = 1
8--Dividing through by 3 yields C - B - B/3 = 1/3
9--Again, ((B + 1)/3 must be an integer k making B = 3k - 1
10--Substitiuing back into (7) yields 3C - 12k + 4 = 1 or C = 4k - 1
11--k.....1.....2.....3.....4.....5.....6.....7
.....B.....2.....5.....8....11...14...17...20
.....C.....3.....7....11...15...19...23...27
12--Combining the 3rd and 4th expressions, we get 2D - 3C = 1
13--Dividing through by 2 yields D - C - C/2 = 1/2
14--Again, (C + 1)/2 must be an integer k making C = 2k - 1
15--Substituting back into (12) yields 2D - 6k + 3 = 1 or D = 3k - 1
16--k....1....2....3....4....5....6....7.....8....9....10....11
.....C....1....3....5....7....9...11..13....15..17...19....21
.....D....2....5....8...11..14...17..20...23..26...29....32
17--Combining the 4th and 1st expressions, we get 2D - 5A = 3
18--Dividing through by 2 yields D - 2A - A/2 = 1 + 1/2
19--Again, (A + 1)/2 must be an integer k making A = 2k - 1
20--Substituting back into (17) yields 2D - 10k + 5 = 3 or D = 5k - 1
21--k....1....2....3....4....5....6....7....8
.....D....4....9...14..19..24...29..34..39
.....A....1....3....5....7....9...11..13..15
22--The smallest consistant set of values throughout the four tables are the highlightd values of A = 11, B = 14, C = 19 and D = 29.
23--Therefore, N = 5(11) + 3 = 58 = 4(14) + 2 = 58 = 3(19) + 1 = 58 = 2(29) = 58.
There are other methods.