If 683+883 is divided by 49 , then the remainder is

Question

If 683+883 is divided by 49 , then the remainder is (A) 35 (B) 5 (C) 1 (D) 0

Tardigrade Admin · Accepted Answer

(1+7)83+(7-1)83=(1+7)83-(1-7)83 =2[ 83 C 1 ⋅ 7+ 83 C 3 ⋅ 73+⋅s+ 83 C 83 ⋅ 783]=(2 ⋅ 7 ⋅ 83)+49 I where I is an integer Now, 14 × 83=1162 (1162/49)=23 (35/49) Therefore, remainder is 35 .

Q. If $6^{83} + 8^{83}$ is divided by $49$ , then the remainder is

35

5

1

0

$(1 + 7)^{83} + (7 - 1)^{83} = (1 + 7)^{83} - (1 - 7)^{83}$
$= 2 [^{83} C_{1} \cdot 7 +^{83} C_{3} \cdot 7^{3} + \dots +^{83} C_{83} \cdot 7^{83}] = (2 \cdot 7 \cdot 83) + 49 I$
where $I$ is an integer
Now, $14 \times 83 = 1162$
$\frac{1162}{49} = 23 \frac{35}{49}$
Therefore, remainder is $35.$

Q. If 683+883 is divided by 49 , then the remainder is

35

5

1

0

(1+7)83+(7−1)83=(1+7)83−(1−7)83 =2[83C1​⋅7+83C3​⋅73+⋯+83C83​⋅783]=(2⋅7⋅83)+49I where I is an integer Now, 14×83=1162 491162​=234935​ Therefore, remainder is 35.

Q. If $6^{83} + 8^{83}$ is divided by $49$ , then the remainder is

$(1 + 7)^{83} + (7 - 1)^{83} = (1 + 7)^{83} - (1 - 7)^{83}$
$= 2 [^{83} C_{1} \cdot 7 +^{83} C_{3} \cdot 7^{3} + \dots +^{83} C_{83} \cdot 7^{83}] = (2 \cdot 7 \cdot 83) + 49 I$
where $I$ is an integer
Now, $14 \times 83 = 1162$
$\frac{1162}{49} = 23 \frac{35}{49}$
Therefore, remainder is $35.$