How to solve CAT 2024 Slot 1 - Quant Question 6?

This question can be solved by applying the core concept explained in the solution above, followed by step-by-step calculation and elimination of options.

Which concept is tested in this question?

This question tests fundamental concepts commonly asked in CAT exams.

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CAT 2024 - Slot 1 Quant Question 6

This is a detailed step-by-step solution for CAT 2024 - Slot 1 Quant Question 6. Understand the concept, common mistakes, and expert tips to solve similar questions in CAT exam.

CAT 2024 Slot 1 - Quant

$\begin{array}{ll}{Rem\, \lgroup{\frac{10}{7}}\rgroup=3} \\ {\Rightarrow Rem\lgroup{\frac{{10}^{100}}{7}}\rgroup=Rem{\lgroup{\frac{10}{7}}\rgroup}^{100}={3}^{100}}\, \, \\ \, \end{array}$ $\begin{array}{ll}{Rem\lgroup{\frac{{3}^3}{7}}\rgroup=(-1)} \\ {Now\, {3}^{100}}\, can\, be\, written\, as\, {3}^{99}\, \times{3}^1={{3}^{33}}^{\times3}\times{3}^1 \\ Thus\, Rem\lgroup{\frac{{3}^{100}}{7}}\rgroup=Rem\lgroup{\frac{{3}^{99}\times{3}^1}{7}}\rgroup=Rem{\lgroup{\frac{\, {{3}^3}^{}{}^{}}{7}}\rgroup}^{33}\times Rem\lgroup{\frac{3}{7}}\rgroup=\lgroup{-1}\rgroup\times\lgroup{3}\rgroup=-3\, \\ \, \end{array}$
Thus remainder = 7 - 3 = 4

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