Aptitude Discussion

Q. |
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? |

✔ A. |
8 |

✖ B. |
9 |

✖ C. |
12 |

✖ D. |
17 |

**Solution:**

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.**

**Chalam Tirunagari**

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**Vijaya**

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

**Yehia**

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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.

**Mansi**

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

kindly explain!

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

$37*2=74$

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

correct it if am wrong

**Misbah Hassan**

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

Misbah,

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:

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

$= \textbf{8}$

Hope that helps

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

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

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

$N_1+N_2 =82$

$\dfrac{82}{37}=8$