Question

If j, k, and n are consecutive integers such that 0 < j < k < n and the units (ones) digit of the product jn is 9, what is the units digit of k?

a. 0
b. 1
c. 2
d. 3
e. 4

Answer 1

it would be good to start with 3

Answer 2

i'm confused with units digit?? what do they mean by that

Answer 3

hmmm...
so we have
j=i-1
k=i
n=i+1
where i is integer
jn=i^2-1
and
jn has number representation:
\[c_n10^{n}+c_{n-1}10^{n-1}+c_{n-2}10^{n-2}+\cdots+c_310^3+c_210^2+c_110^1+c_010^0\]
and we given c_0 is 9...
...
still thinking on how to proceed with this

Answer 4

most likely, the units here are referred to the place value
123, to hundreds
12, tens
1, ones

Answer 5

0.1 tenth
0.34 hundreth

etc.

Answer 6

http://www.mathsisfun.com/place-value.html

Answer 7

i know the answer is a, i just don't know how they got it

Answer 8

it cannot be a since 0 < j
0 = 0 and not 0 < 0

Answer 9

\[i^2-1=.....9 \\ \]
what is the unit digit for i^2 if you all 1 on both sides

Answer 10

that's what the answer key says though, nin

Answer 11

oh LAUGHING OUT LOUD well then

Answer 12

j>0
doesn't mean the units digit of j is 0

Answer 13

j could be 10 for all we know
10 is still bigger than 0

Answer 14

I think the easier way it to add 9 on both sides of i^2-1=.....9

Answer 15

i'm like, really confused right now

Answer 16

so how do we satisfy the product jn = 9

Answer 17

and say that i^2 has unit digit?

Answer 18

and j < n

Answer 19

i^2-1=.....9
i^2=.......0
so i has to to have unit digit?

Answer 20

i will let you finish it from here

Answer 21

I do not see why j = 0
maybe because I just woke up. I will think about this, but right now sorry that I am not of any help

Answer 22

freckles, i don't know what to do ._.

Answer 23

first do you understand what i have written

Answer 24

no :x

Answer 25

he is suggesting that the values j and k are not straight-forward so he is indicating them as i

j = i - 1
k = i
n = i + 1
which has to satisfies j < k <n
but our condition is 0 < j < k < n
this means that our j cannot be 0

Answer 26

jn is i^-1 and jn has one vale 9

Answer 27

i^2-1*

Answer 28

is that i^{2-1} ?

Answer 29

so jn looks like some numbers then 9

Answer 30

no (i-1)(i+1)=i^2-1

Answer 31

i^2-1=......9

Answer 32

are you referring to THE imaginary number @freckles ?

Answer 33

ok i get that part now

Answer 34

those dots just represent the digits in the 10s 100s and so on place value

Answer 35

lol no
i said i was an integer above

Answer 36

then use Z next time LAUGHING OUT LOUD

Answer 37

are you purposely giving me a hard time :p

Answer 38

look i defined i above

Answer 39

in my first post

Answer 40

freckles, i get you lol. what we do from here?

Answer 41

what do we do*

Answer 42

we have i^2-1=.......9

Answer 43

if you add 1 to both sides that changes the unit digit only for the right hand side

Answer 44

after all 1+9=10
so you i^2=.......0

Answer 45

so i=.......0

Answer 46

\(\large\tt \begin{align} \color{black}{0<j<k<n\\~\\
\implies 0<j<j+1<j+2\\~\\
(j\times n )mod 10\equiv 9\\~\\
(j\times (j+2) )mod 10\equiv 9\\~\\
j=19\\~\\
0<j<k<n\\~\\
\implies 0<19<20<21\\~\\
j\times n=19\times 21=399\\
\text{it satisfies the condition}\\~\\
\text{units digit of k is zero}\\
}\end{align}\)

Answer 47

wait whaaaat

Answer 48

i still don't get it. i guess i'm stupid

Answer 49

u can use the modular arthmatic method if u know it

Answer 50

i do not know it :/

Answer 51

the only integer number i can square and get 0 as a unit digit is a number that also has unit value 0
right?
0^2=0
1^2=1
2^2=4
3^2=9
4^2=16
5^2=25
6^2=36
...
10^2=100

Answer 52

ok

Answer 53

so if i^2=.......0
then i=..........0

Answer 54

but how do you get the answer of 0?

Answer 55

we just did

Answer 56

recall i is k

Answer 57

oh wait i think i get it now. so because i^2 - 1 = ....9 and adding 1 to both sides would be i^2 = .....0 and you can only get 0 in the unit place when you square a number, only if the number itself ends in 0, then the answer must be 0

Answer 58

yeah

Answer 59

ok thank you very much :)