Solveeit Logo
Login

Question

Question: Find the remainder, if \(7^{103}\) is divided by 25....

Find the remainder, if 71037^{103} is divided by 25.

Explanation

Solution

Hint: In this question it is given that if we divide 71037^{103} by 25 then what will be the remainder. So to prove this we need to know the binomial expansion, which is-
(a−b)n= nC0 an− nC1 an−1b+ nC2 an−2b2−⋯+(−1)n nCn bn\left( a-b\right)^{n} =\ ^{n} C_{0}\ a^{n}-\ ^{n} C_{1}\ a^{n-1}b+\ ^{n} C_{2}\ a^{n-2}b^{2}-\cdots +\left( -1\right)^{n} \ ^{n} C_{n}\ b^{n}......(1)
And after that if we divide 71037^{103} by 25 then this will give us the reminder.
Complete step-by-step solution:
Given term can be written as,
7103=7102+17^{103}=7^{102+1}
=7⋅7102=7\cdot 7^{102} [since, am+n=am⋅ana^{m+n}=a^{m}\cdot a^{n}]
=7(72)51=7\left( 7^{2}\right)^{51}
=7(49)51=7\left( 49\right)^{51}
=7(50−1)51=7\left( 50-1\right)^{51}
Now we are going to expand the binomial (50−1)51\left( 50-1\right)^{51} where a = 50, b = 1 and n = 51, we get,
7(50−1)517\left( 50-1\right)^{51}
=7[ 51C0 5051− 51C1 5050⋅1+ 51C2 5049⋅12−⋯ + 51C50 50⋅150− 51C51151]=7[\ ^{51} C_{0}\ 50^{51}-\ ^{51} C_{1}\ 50^{50}\cdot 1+\ ^{51} C_{2}\ 50^{49}\cdot 1^{2}-\cdots \ +\ ^{51} C_{50}\ 50\cdot 1^{50}-\ ^{51} C_{51}1^{51}]
=7[5051− 51C1 5050+ 51C2 5049−⋯ + 51C50 50− 1]=7[50^{51}-\ ^{51} C_{1}\ 50^{50}+\ ^{51} C_{2}\ 50^{49}-\cdots \ +\ ^{51} C_{50}\ 50-\ 1] [
=7⋅5051−7⋅ 51C1 5050+7⋅ 51C2 5049−⋯ +7⋅ 51C1 50− 7=7\cdot 50^{51}-7\cdot \ ^{51} C_{1}\ 50^{50}+7\cdot \ ^{51} C_{2}\ 50^{49}-\cdots \ +7\cdot \ ^{51} C_{1}\ 50-\ 7
=7⋅5051−7⋅ 51C1 5050+7⋅ 51C2 5049−⋯ +7⋅ 51C1 50− 7 −18 +18 =7\cdot 50^{51}-7\cdot \ ^{51} C_{1}\ 50^{50}+7\cdot \ ^{51} C_{2}\ 50^{49}-\cdots \ +7\cdot \ ^{51} C_{1}\ 50-\ 7\ -18\ +18\ [by adding and subtracting 18]
=7⋅5051−7⋅ 51C1 5050+7⋅ 51C2 5049−⋯ +7⋅ 51C1 50− 25+18=7\cdot 50^{51}-7\cdot \ ^{51} C_{1}\ 50^{50}+7\cdot \ ^{51} C_{2}\ 50^{49}-\cdots \ +7\cdot \ ^{51} C_{1}\ 50-\ 25+18
Apart from the last term, in each and every term 50 and 25 is there which is the multiple of 25 so we can write the first (n+1) terms as 25k,
i.e, 25k=7⋅5051−7⋅ 51C1 5050+7⋅ 51C2 5049−⋯ +7⋅ 51C1 50− 25 25k=7\cdot 50^{51}-7\cdot \ ^{51} C_{1}\ 50^{50}+7\cdot \ ^{51} C_{2}\ 50^{49}-\cdots \ +7\cdot \ ^{51} C_{1}\ 50-\ 25\
Therefore,
7103=25k+187^{103}=25k+18
So we can say that after dividing 25, we get the reminder 18.
Note: While solving this type of question you need to know that when you divide any term or any polynomial then the obtained reminder is always positive, that is why we add and subtract 18 to convert the term -7 into -25, i.e, the multiple of 25, which gives the reminder 18.