Question

the 3 numbers 1428, 2134, and 2850 has the same remainder y when they all divided by x. the possible value of x + y = ....

Answer 1

@ganeshie8

Answer 2

Is this what the values mean:
\[y=\frac{1428}{x}\]\[y=\frac{2134}{x}\]\[y=\frac{2850 }{x}\]

Answer 3

px + y = 1428
qx + y = 2134
rx + y = 2850

Answer 4

In other words,
\[1428\equiv2134\equiv2850\equiv x\mod y\,?\]

Answer 5

Sorry, should be \(\color{red}{y\mod x}\).

Answer 6

y its remainder for that numbers

Answer 7

yes, y modx
what's next ?

Answer 8

Notice that
\[\begin{cases}2850-2134\equiv y-y\mod x~~\implies~~716\equiv 0\mod x\\
2134-1428\equiv y-y\mod x~~\implies~~706\equiv0\mod x
\end{cases}\]
In turn,
\[716-706\equiv0-0\mod x~~\implies~~10\equiv0\mod x\]

Answer 9

is the answer 2 ?

Answer 10

still didnt understand, what do you means ?

Answer 11

here the choices answer :
a. 165 b. 179 c. 344 d. 716

Answer 12

You can divide 10 by four different integers such that the remainder is 0 (i.e. \(10\equiv0\mod x\)), which suggests \(x=1\), \(2\), \(5\), or \(10\).

Of these four, only \(x=2\) agrees with the fact that \(706\equiv0\mod 2\), and so on with the others. The fact that the three given numbers are all even suggests that \(y=0\).

Answer 13

I'm agreeing with @ganeshie8 but there may be more to this modular arithmetic than what I can gather from glancing at the wiki page on it.

Answer 14

ah, sorry... extra information says with y not equal zero

Answer 15

Haha! now it is interesting

Answer 16

i think if y = 0, the question not enough interesting :p

Answer 17

yes y = 0, x = 2 is a trivial solution..
going for dinner.. il give a good try after coming back if it still remains unsolved

Answer 18

ok, thanks :D

Answer 19

@mathmath333 try this :)

Answer 20

\(\large \color{black}{
1428=px+y\hspace{.33em}\\~\\
2134=qx+y\hspace{.33em}\\~\\
2850=rx+y\hspace{.33em}\\~\\\\~\\\\~\\
\normalsize \text{substracting those}\hspace{.33em}\\~\\\\~\\\\~\\
706=(q-p)x\hspace{.33em}\\~\\
716=(r-q)x\hspace{.33em}\\~\\
1422=(r-p)x\hspace{.33em}\\~\\~\\~\\
\color{red}{2}\times 353=(q-p)\color{red}{x}\hspace{.33em}\\~\\
\color{red}{2}\times2\times 179=(r-q)\color{red}{x}\hspace{.33em}\\~\\
\color{red}{2}\times 3^2\times79 =(r-p)\color{red}{x}\hspace{.33em}\\~\\\\~\\\\~\\
\normalsize \text{as gcd of those is 2,only x=2 makes sense}\hspace{.33em}\\~\\
\normalsize \text{which is the largest number which leaves the same remainder}\hspace{.33em}\\~\\
\normalsize \text{but when x=2 ,y=0 , and u dont want that condition.}\hspace{.33em}\\~\\
\normalsize \text{so i think may be something is wrong with that numbers may be a typo}\hspace{.33em}\\~\\

}\)

Answer 21

I whipped up a script in Mathematica to check the remainders for values of \(x\) up to \(100,000\) (unnecessarily high, I know) that suggests \(x=1\) or \(x=2\), with \(y=0\) in both cases. I agree with @mathmath333, there must be a problem with the given answers or the question.