The remainder, when 19 200 + 23 200 is divided by 49, is ________ . [2023]

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
The remainder, when $19^{200} + 23^{200}$ is divided by 49, is ________ . [2023]

Ans.
(29)

$23^{200} + 19^{200} = {(21 + 2)}^{200} + {(21 - 2)}^{200}$

$= 2 [{}^{200}{C_{0}} 21^{200} 2^{0} + {}^{200}{C_{2}} 21^{198} 2^{2} + \dots + {}^{200}{C_{200}} 21^{0} 2^{200}]$ ...(i)

Now, $21^{2}$ will be divisible by 49, so (i) can be written as $2 [49 λ + 2^{200}]$

$\frac{2 [49 λ + 2^{200}]}{49}$ $then, Remainder = \frac{2^{201}}{49} = \frac{{(8)}^{67}}{49} = \frac{{(7 + 1)}^{67}}{49}$

$= \frac{{}^{67}{C_{0}} 7^{67} \cdot 1^{0} + {}^{67}{C_{1}} 7^{66} \cdot 1 + \dots + {}^{67}{C_{66}} 7 \cdot 1^{66} + {}^{67}{C_{67} \cdot 1}}{49}$

$Remainder = \frac{67 \times 7 + 1}{49}$ $[other terms will be divisible by 49]$

$= \frac{(49 + 18) \times 7 + 1}{49}, Remainder = 29$

Want Us to Email you About Special Offers & Updates?