JEE Main 2023 — Mathematics Algebra

Using the binomial expansion in ${7}^{103}={({7}^{2})}^{51}\cdot 7={(51-2)}^{51}\cdot 7$7103=(72)51⋅7=(51−2)51⋅7 we get,

$${(51-2)}^{51}\cdot 7=7\cdot (C051{51}^{51}\cdot {2}^{0}-C151{51}^{50}\cdot {2}^{1}+.......-C5151{2}^{51})$$(51−2)51⋅7=7⋅(C0515151⋅20−C1515150⋅21+.......−C5151251)

Now $51$51 is divisible by $17$17, so when above equation is divided by $17$17 we get, $7\cdot (-{2}^{51})$7⋅(−251) as remainder,

Now again using binomial in $7\cdot (-{2}^{51})=-56{({2}^{4})}^{12}=-56{(17-1)}^{12}$7⋅(−251)=−56(24)12=−56(17−1)12 we get,

$$-56{(17-1)}^{12}=-56(C012{17}^{12}\cdot {1}^{0}-C112{17}^{11}\cdot {1}^{1}+.......+C1212{1}^{12})$$−56(17−1)12=−56(C0121712⋅10−C1121711⋅11+.......+C1212112)

Now dividing above equation by $17$17 we get,

$(-56\times 1)\div 17=-5$(−56×1)÷17=−5,

Now changing negative remainder to positive we get, $-5+17=12$−5+17=12,

Hence, when ${7}^{103}$7103 divided by $17$17 gives $12$12 as remainder.