The remainder, when 7^ 103 is divided by 23 , is equal to: - Vidyadip

MCQ

The remainder, when $7^{103}$ is divided by 23 , is equal to:

✓
14
B
9
C
17
D
6

✓

Answer

Correct option: A.

14

(A) 14
$
\begin{array}{l}Sol.\
7^{103}=7\left(7^{102}\right)=7(343)^{44}=7(345-2)^{34} \\
7^{103}=23 K_1+7.2^{44} \\
\text { Now } 7.2^{34}=7 \cdot 2^2 \cdot 2^{12} \\
=28 \cdot(256)^4 \\
=28(253+3)^4 \\
\therefore 28 \times 81 \Rightarrow(23+5)(69+12) \\
23 K_2+60 \\
\therefore \text { Remainder }=14
\end{array}
$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "SECTION - A [MATHS - MCQ]" from JEE Main 29-Jan-2025 Paper - Shift 2→JEE Main 29-Jan-2025 Paper - Shift 2 — all question types→JEE STD 11 Science question papers→Gujarat Board STD 11 Science question papers→Gujarat Board question paper generator→

Similar questions

On the sides $AB, BC, CA$ of a $\Delta ABC, 3, 4, 5$ distinct points ( excluding vertices $A, B, C$) are respectively chosen. The number of triangles that can be constructed using these chosen points as vertices are

If $arg\,z < 0$ then $arg\,( - z) - arg\,(z)$ is equal to

The value of ${1^2} + {3^2} + {5^2} + ....... + {25^2}$ is

$\sum\limits_{n = 1}^\infty {\sum\limits_{k = 1}^{n - 1} {\frac{k}{{{2^{n + k}}}}} } $ is equal to

For a scholarship, atmost $n$ candidates out of $2n+1$ can be selected. If the number of different ways of selection of atleast one candidate for scholarship is $63$, then maximum number of candidates that can be selected for the scholarship is -

Let $y=\log _8\left(\frac{1-x^2}{1+x^2}\right),-1$$-1 < x< 1 $ Then at $ x=\frac{1}{2},$ the value of $225\left(y^{\prime}-y^{\prime \prime}\right)$ is equal to

Let $a, a r, a r^2, \ldots . . .$. be an infinite $G.P.$ If $\sum_{n=0}^{\infty} a^n=57$ and $\sum_{n=0}^{\infty} a^3 r^{3 n}=9747$, then $a+18 r$ is equal to :

Let the coefficients of the middle terms in the expansion of $\left(\frac{1}{\sqrt{6}}+\beta x\right)^{4},(1-3 \beta x)^{2}$ and $\left(1-\frac{\beta}{2} x\right)^{6}, \beta>0$, respectively form the first three terms of an $A.P.$ If $d$ is the common difference of this $A.P.$, then $50-\frac{2 d}{\beta^{2}}$ is equal to.

A bag contains $5$ distinct Red, $4$ distinct Green and $3$ distinct Black balls. Balls are drawn one by one without replacement,then the probability of getting a particular red ball in fourth draw is-

Numbers are to be formed between $1000$ and $3000$, which are divisible by $4$, using the digits $1,2,3,4,5$ and $6$ without repetition of digits. Then the total number of such numbers is.