Question

For a nonnegative integer n, let r_9(n)stand for the remainder left when n is divided by 9. For example, r_9(25)=7.

What is the 22nd entry in an ordered list of all nonnegative integers n that satisfy r_9(5n) =<4?
(Note that the first entry in this list is 0.)

Answer 1

n = 0 r_9(5*0) = 0
n = 1 r_9(5*1) = 5
n = 2 r_9(5*2) = 1
n = 3 r_9(5*3) = 6

so the first 2 are 0, 2

Answer 2

n = 4 would give 2
n = 5 7
n = 6 3

Answer 3

@welshfella
0, 2, 4, 6... that's a pattern by 2's

Answer 4

yea looks that way

Answer 5

@welshfella does that mean the answer is 42?

Answer 6

seems ok
when n = 42 r_9(5*42) = 3

Answer 7

@welshfella Alcumus says it's incorrect....

Answer 8

oh ok

lets try some more
n = 7 gives 35/9 rem 8
n = 8 gives 40 / 9 rem 4

so n = 8 is not on the list

Answer 9

n = 9 gives 45/9 rem 0 so n=9 is included in the list

so thats why we are wrong.

the first five are are 0,2,4,6, 9

Answer 10

the next one is n = 11 because 55/9 gives rem of 1

Answer 11

n=13 gives rem 2
n = 15 gives rem 3

so they are going up in odd numbers now

Answer 12

so..
0, 2, 4, 6, 9, 11, 13, 15......
how do you know when it change? like a pattern to find the 22nd number

Answer 13

try some more values
16,17 18 ,19
what do we get?

this is a tiresome way to answer the question but I cant think of another one at the moment

Answer 14

16 not included
17 -- 5*17 = 85 85/9 - remainder 4 so 17 is not included

Answer 15

- maybe we'll get even numbers now....
n=18 5*18 = 90 - rem 0 so n = 18
n=19 - nope
n=20 5820 / 9 rem 1 so n = 20

Answer 16

0,2,4,6,9,11,13,15, 18,20,22,24

Answer 17

maybe they are in groups of 4 alternate even and odd

Answer 18

if this is so then the next num ber should be n = 27

n = 25 5*25 = 125 / 9 rem 8
n = 26 5*26 = 130/9 rem 4
n = 27 135 135 /9 rem 0

looks that way
after the evens/odds theres a jump of 3

Answer 19

but 26 would also work because the remainder can equal 4... its =<

Answer 20

no 26 not included

Answer 21

$$r_9(5n)\le 4~?$$

Answer 22

but 26 should work bc remainder 4..^

Answer 23

the next 4 would be 27 29 31 33

then 36 18 40 42

Answer 24

but the remainder can equal 4 also

Answer 25

Oh i misread the question I took it as < 4

Answer 26

It's alright :)

i think the only change is that it's groups of 5 and a jump of 2?

Answer 27

0, 2, 4, 6, 8, 9, 11, 13, 15, 17, 18, 20, 22, 24, 26, 27, 29, 31, 33, 35, 36, 38, 40, 42, 44

groups of 5 and jump of 1

Answer 28

yea n = 8 has to be included
also n = 17 and n = 26

Answer 29

sothe 22nd would be 38

Answer 30

yup its correct :)

answer alcumus gave:

The condition $r_9(5n)\le 4$ can also be stated as $``5n\equiv 0,1,2,3,\text{ or }4\pmod 9."$'

We can then restate that condition again by multiplying both sides by $2:$ $$10n \equiv 0,2,4,6,\text{ or }8\pmod 9.$$This step is reversible (since $2$ has an inverse modulo $9$). Thus, it neither creates nor removes solutions. Moreover, the left side reduces to $n$ modulo $9,$ giving us the precise solution set $$n \equiv 0,2,4,6,\text{ or }8\pmod 9.$$We wish to determine the $22^{\text{nd}}$ nonnegative integer in this solution set. The first few solutions follow this pattern: $$\begin{array}{c c c c c}
0 & 2 & 4 & 6 & 8 \\
9 & 11 & 13 & 15 & 17 \\
18 & 20 & 22 & 24 & 26 \\
27 & 29 & 31 & 33 & 35 \\
36 & 38 & \cdots
\end{array}$$The $22^{\text{nd}}$ solution is $\boxed{38}.$

Answer 31

thanks so much for all ofyour help they did it basically the same way except mods ;)

@welshfella

Answer 32

thats correct

Answer 33

yw