Show that one and only one out of n,n+4,n+8,n+12,n+16 is divisible by 5,where n is any positive integer. A two digit number is obtained by either multiplying the sum of the digits by 8 and then subtracting 5 or by multiplying the differences of the digits by 16 and then adding 3.help me immediately.
thhhhhhhhiiiiiiiiiiiisssssssss sssssssuuuuuummmmmmm iiiiiissss tttoooooughh
To show that only one out of the numbers n, n+4, n+8, n+12, n+16 is divisible by 5, we need to consider the properties of numbers divisible by 5.
A number is divisible by 5 if and only if its units digit is either 0 or 5.
Let's analyze the given numbers one by one:
1. n: Since n is any positive integer, we cannot make any assumptions about its units digit.
2. n+4: Adding 4 to n will not change the units digit, so it will have the same divisibility properties as n.
3. n+8: Adding 8 to n will not change the units digit, so it will have the same divisibility properties as n.
4. n+12: Adding 12 to n can potentially change the units digit. Here, we can observe that when the units digit of n is 3, 8, or 13, the units digit of n+12 will be 5. Thus, n+12 is divisible by 5 for some values of n.
5. n+16: Adding 16 to n can also potentially change the units digit. Similarly, we can observe that when the units digit of n is 9, 14, or 19, the units digit of n+16 will be 5. Thus, n+16 is divisible by 5 for some values of n.
From the above analysis, we can conclude that out of the given numbers, only one of n+12 and n+16 is divisible by 5. This means that if n is any positive integer, then one and only one of the numbers n, n+4, n+8, n+12, n+16 is divisible by 5.
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