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A certain number when divided by 222 leaves a remainder 35, another number when divided by 407 leaves a remainder 47.

What is the remainder when the sum of these two numbers is divided by 37?









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Option(A) is correct

$N_1 = 222x+35$ and $N_2 = 407y+47$

$N_1 + N_2 = (37×6×x+35) + (37×11×y +47)$

Remainder when $N_1 + N_2$ is divided by $37$,

$= \dfrac{35+47}{37} = 8$

Edit: For an alternative solution, check comment by Som.

Edit 2: For yet another alternative solution, check comment by Yehia.

Edit 3: For yet-yet another alternative solution, check comment by Chalam Tirunagari.

(9) Comment(s)

Chalam Tirunagari

$N_1=222x+35$ and $N_2=407y+47$

Since two eqations are independent take $x=0$ $y=0$

Then $N_1=35$, $N_2=47$

$N_1+N_2 =82$



$=\dfrac{(37×6×x+35)+(37×11×y+47)}{37}$ is clear.

Now how do we get, $\dfrac{37+47}{37}$?


You have to get $= \dfrac{35+47}{37}$ and not $= \dfrac{37+47}{37}$

Here goes the part explaining this.

Since the remainder is to be found, when $\dfrac{(37×6×x+35) + (37×11×y +47)}{37}$ is looked at, terms $37×6×x$ and $37×11×y$ leave no remainder (or remainder is '0') so they are dropped.

We are only left with $\dfrac{35+47}{37}$ which leaves the remainder of $8$.

And that is the final answer.


Let $\dfrac{m}{222} = x$ then $m= 222x+35$

Take $x=1$ then $m= 222+35= 257$

Let $\dfrac{n}{407}= y$ then $n=407y+47$.

Take any value for $y (y=1)$ then

Then $n= 407+ 47= 454$

Now $m+ n = 257+454=711$

$\dfrac{711}{37}= 19$ with the remainder 8.

Trick is to plug in numbers to make it easier. This is the beauty of multiple choice question. The question does not ask u to prove anything.

All u need is to find the right answer.


I didn't get the last part.How we get $\dfrac{37+47}{37}$?...

kindly explain!


please explain how $\dfrac{37+47}{37}=\dfrac{84}{37} =8$???


$84-74=$ leaves the reminder 10 not 8 please

correct it if am wrong

Misbah Hassan

Kindly explain this question. i am not able to solve this question



Its easy, lets assume that two numbers be $N_1$ and $N_2$. Based on the given information $N_1=222x+35$ and $N_2=407y+47$

Now sum of these,

$N_1 + N_2 = (222x+35)+ (407y+47)$

We have to find out the remainder when it is divided by 37.

So, $\dfrac{N_1 +N_2}{37}$

$=\dfrac{(222x+35)+ (407y+47)}{37}$

$=\dfrac{(37\times 6\times x+35)+(37\times 11\times y+47)}{37}$

So, Remainder when $N_1+N_2$ is divided by 37:


$= \textbf{8}$

Hope that helps Smile