The remainder when 22003 is divided by 17 is

Question

The remainder when 22003 is divided by 17 is (A) 7 (B) 8 (C) 2 (D) 1

Tardigrade Admin · Accepted Answer

We have, 22000=(24)500=(17-1)500 = 500 C0 17500- 500 C1 17499+ ldots- 500 C499 17+(-1)500 17 m+1, where m is some positive integer =22003=(8)(22000)=8(17 m+1) =17(8 m)+8 Hence, the remainder will be 8 .

Q. The remainder when $2^{2003}$ is divided by 17 is

7

8

2

1

0

We have,
$2^{2000} = (2^{4})^{500} = (17 - 1)^{500}$
$=^{500} C_{0} 1 7^{500} -^{500} C_{1} 1 7^{499} + \dots -^{500} C_{499} 17 + (- 1)^{500}$
$17 m + 1$ , where $m$ is some positive integer
$= 2^{2003} = (8) (2^{2000}) = 8 (17 m + 1)$
$= 17 (8 m) + 8$
Hence, the remainder will be $8$ .

Q. The remainder when 22003 is divided by 17 is

7

8

2

1

0

We have, 22000=(24)500=(17−1)500 =500C0​17500−500C1​17499+…−500C499​17+(−1)500 17m+1, where m is some positive integer =22003=(8)(22000)=8(17m+1) =17(8m)+8 Hence, the remainder will be 8 .

Q. The remainder when $2^{2003}$ is divided by 17 is

We have,
$2^{2000} = (2^{4})^{500} = (17 - 1)^{500}$
$=^{500} C_{0} 1 7^{500} -^{500} C_{1} 1 7^{499} + \dots -^{500} C_{499} 17 + (- 1)^{500}$
$17 m + 1$ , where $m$ is some positive integer
$= 2^{2003} = (8) (2^{2000}) = 8 (17 m + 1)$
$= 17 (8 m) + 8$
Hence, the remainder will be $8$ .