Question

qid : 8 - n is the largest two digit number, which when divided by 3, 4 and 6 leaves the remainder 1, 2 and 4 respectively. what is the remainder when n is divided by 5?

Answer 1

I got N=94, remainder 4, same as others. I solved it this way:

x,y,z are unknown integers, and N<100 (two digits)

N = 3x + 1, N = 4y + 2, and N = 6z + 4 must all be true simultaneously, based on the remainder statements. N must be even based on the second two statements.

Also, if N = 6z + 4 < 100 then z<16

The second two statements I rewrote:

N/2 = 2y + 1, therefore N/2 is odd

N/2 = 3z + 2 is odd, therefore z must be odd. The allowed values of z<16 are:

1, 3, 5, 7, 9, 11, 13, 15

Since we are asked for the largest possible N, this corresponds to the largest z=15.

N = 6*15 + 4 = 94, with remainder 4 if divided by 5.

This solution also corresponds to x = 31, y =23.

Additional note: In general for any N<100, x>y>z and the smallest solution that satisfies the remainder condition is:

N = 10, x=3, y=2, z=1 and remainder zero if divided by 5 (as mentioned in another answer).

All of the ordered triples ( x, y, z ) that satisfy the three remainder conditions are:

(3, 2, 1), (7, 5, 3) , (11, 8, 5), (15, 11, 7), (19, 14, 9), (23, 17, 11), (27, 20, 13), (31, 23, 15)

with N = 10, 22, 34, 46, 58, 70, 82, 94 respectively.

with remainders mod 5: 0, 2, 4, 1, 3, 0, 2, 4

The N values are evenly spaced 12 apart and the remainder sequence repeats as we go above N = 99.