Posted by Abhishek kumar sahi on Thursday, August 4, 2011 at 10:28am.
Hints for first part:
0 mod 5 = 0
4 mod 5 = 4
8 mod 5 = 3
12 mod 5 = 2
16 mod 5 = 1
Use (a+b)mod 5 = (a mod 5 + b mod 5)mod 5.
write the sequence as
(n) + 0 , (n)+4, (n+5) + 3 , (n+10) + 2, (n+15) + 1
dividing any number n by 5 will leave remainder of 1, 2, 3, 4, or 0 (a remainder of zero will give an exact divsion)
The + ? at the end would be the possible remainders for different values of n, and all possible remainders show up.
e.g
if n = 1, the 2nd, or (n) + 4 is divisible by 5
if n = 2 , the 3rd or (n+5)+3 is divisible by 5
if n = 3, the 4th, or (n+10) + 2 is divisible by 5
if n = 4 , the 5th, or (n+15) + 1 is divisible by 5
if n = 5 , 1st, or n is divisible by 5
if n = 6 , the 2nd, or (n) + 4 is divisible by 5
and the cycle keeps going like that.
Notice that for any value of n, there is only one of the expressions that would be divisible by 5
Your second question:
let the unit digit be x, let the tens digit by y
the number would be 10y+x
8(x+y) - 5 = 16(x-y) + 3
8x + 8y - 5 = 16x - 16y + 3
8x - 24y = -8
x = 3y - 1
let y = 1 , x = 2 , number is 12
let y = 2, x = 5 , number is 25
let y = 3, x = 8 , number is 38
test:
for 12, is 8(1+2)- 5 = 16(2-1) + 3 ? Yes
for 25, is 8(2+5) - 5 = 16(5-2) +3 ? yes
for 38, is 8(3+8) - 5 = 16(8-3) + 3 ? Yes
the two digit number could be 12, 25, or 38