Question

The number n is a 2-digit number. When n is divided by 10, the remainder is 9, and when n is divided by 9, the remainder is 8. What is the value of n?

Answer 1

hINT :
\(dividend = divisor * quotient+remainder.\)

Answer 2

PLEASE dont give the written answer PEOPLE!

Answer 3

do you want a clue? @Areesha.1D

Answer 4

yeah @hartnn i think provided it? one minute, let me try.

Answer 5

I did it like this, n/10=x+9 and n/9=x+8 and then found n in both and solve for X. Where did i go wrong?

Answer 6

well another way of solving is using the modulus. Are you familiar with it?
let n=10a +b,
\(10a+b \equiv 9 (mod 10)\)
\(10a+b \equiv 8 (mod 9)\)
note that \(a \equiv b (mod n)\) then \(a-b \equiv 0 (mod n)\)

Answer 7

if you divide 23 or 43 by 10, the remainder is ___ ?

Answer 8

u were incorrect in taking both x's.....

Answer 9

@Shadowys yeah i don't....

Answer 10

@hartnn i know, but i was trying to find a way... trial and error ya know?

Answer 11

let's narrow down the choices. if you divide 23 or 43 by 10, the remainder is 3. So, what kind of number leaves a remainder of 9 when diided by 10?

Answer 12

in the language of fractions, \[\large \frac{ 2\color{red}3 }{ 10 }=2\frac{ \color{red}3 }{ 10 }\\ \large \frac{ 4\color{red}3 }{ 10 }=2\frac{ \color{red}3 }{ 10 }\]

Answer 13

well, it's quite handy in this question.
\(10a+b≡9(mod10)\)
\(10a+b-9≡0(mod10)\) means that \(\frac{10a+b-9}{10} =0\)
so, \(a+ \frac{b-9}{10} =0\), so since a and b are integers<10....I guess b is known... well, then do the same thing with mod 9.

Answer 14

49 29 etc

Answer 15

@Shadowys solution reveals that the last digit is 9. so the choices are 19, 29, 39, up to 89

Answer 16

I dont understand shadowys solution!

Answer 17

I do, however, underrstand the last digit has to be 9!!

Answer 18

trial and error should solve the problem. there are a few possible choices.

Answer 19

the mod form is basically a simplified dividend=... formula.
\(a \equiv b (mod n)\) means a/n will have remainder b. This gets rid of solving for (x) and also can show that since remainder is b, then if we take out b from a then it can be divided perfectly.
so \(a-b \equiv 0 (mod n)\)

note that n has two digit, so it is made of one tens and one ones, i.e. it can be written as n=10a+b, where a,b<10, and they are positive integers.

the second case is \(10a+9 \equiv 8 (mod 9)\)
so, \(10a+9-8 \equiv 0 (mod 9)\)
\(10a+1 \equiv 0 (mod 9)\)
the only suitable value for a is....?

Answer 20

you can also say 10a + 1 = a + 1 = 0 (mod 9)

Answer 21

WHAT is mod?

Answer 22

mod is..well, the above expression for dividends and remainders. The explanation is above though.

Answer 23

@Shadowys solution is a foreign language =(

Answer 24

its just he knows modular arithmetic...

Answer 25

why don't you just divide each of the numbers 19, 29, 39, up to 89 and check which leaves a remainder of 8.

Answer 26

i wouls also use trial and error, once i know units digit is 9...just like
@sirm3d said....

Answer 27

the smallest positive number that leaves a remainder of 8 when divided by 9 is 8. agree? @Areesha.1D

Answer 28

lol sorry about those...kinda foreign to me too actually...just got around using it

Answer 29

So trial and error it is. Thanks Guys!!!

Answer 30

@Areesha.1D if you successively add 9 to 8, what remainder will you get from each number after division by 9?

Answer 31

8

Answer 32

good. so which of the numbers END in 9?

Answer 33

19 29 39 etc

Answer 34

division by 9, the numbers are : 8, 17, 26, 35, 44, 53, 62, 71, 80, 89, 98, ...
division by 10, the numbers are : 9, 19, 29, 39, 49, 59, 69, 79, 89, 99, ...

Answer 35

can you identify the common number in the list?

Answer 36

right 89

Answer 37

BINGO.