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Question: What is remainder when \[{7^{84}}\] is divided by \(2402?\) (A) \(1\) (B) \(6\) (C) \(2401\...

What is remainder when 784{7^{84}} is divided by 2402?2402?
(A) 11
(B) 66
(C) 24012401
(D) None of these

Explanation

Solution

The Binomial theorem tells us how to expand expressions of the form (a+b)n{(a + b)^n} . The larger the power is, the harder it is to expand expressions directly. But with the Binomial theorem, the process is relatively easy and fast. First we simplify the given function and then calculate. After simplifying we use the Binomial theorem and get the remainder term.
Formula of expand by using binomial theorem, we get (a+b)n=nC0(a)0×(b)n+n−1C1(a)1×(b)n−1+n−2C2(a)2×(b)n−2+.....+nCn(a)n×(b)0{(a + b)^n}{ = ^n}{C_0}{(a)^0} \times {(b)^n}{ + ^{n - 1}}{C_1}{(a)^1} \times {(b)^{n - 1}}{ + ^{n - 2}}{C_2}{(a)^2} \times {(b)^{n - 2}} + .....{ + ^n}{C_n}{(a)^n} \times {(b)^0}

Complete step-by-step solution:
Given the function 784{7^{84}}
Simplifying the above function and we get
=(74)21= {({7^4})^{21}}
We know that the 74=7×7×7×7{7^4} = 7 \times 7 \times 7 \times 7
=49×49= 49 \times 49
=2401= 2401
Put this in above function and we get
=(2401)21= {(2401)^{21}}
=(2402−1)21= {(2402 - 1)^{21}}
Now we use binomial theorem and expand the above function
=21C0(2402)0×(−1)21+21C1(2402)1×(−1)20+21C2(2402)2×(−1)19+.....+21C21(2402)21×(−1)0{ = ^{21}}{C_0}{(2402)^0} \times {( - 1)^{21}}{ + ^{21}}{C_1}{(2402)^1} \times {( - 1)^{20}}{ + ^{21}}{C_2}{(2402)^2} \times {( - 1)^{19}} + .....{ + ^{21}}{C_{21}}{(2402)^{21}} \times {( - 1)^0}
Find the value of the combination 21C0^{21}{C_0}
Therefore 21C0=21!0!(21−0)!^{21}{C_0} = \dfrac{{21!}}{{0!(21 - 0)!}}
=21!0!×21!= \dfrac{{21!}}{{0! \times 21!}}
=10!= \dfrac{1}{{0!}}
We know that 0!=10! = 1 , use this and we get
=1= 1
Use this 21C0=1^{21}{C_0} = 1 and we get
=−1+21C1×2402+....+21C21(2402)21= - 1{ + ^{21}}{C_1} \times 2402 + ....{ + ^{21}}{C_{21}}{(2402)^{21}}
Except the first term, from second, all the terms are divisible by 24022402
We take all the divisible part as a constant kk , we get
=−1+2402k= - 1 + 2402k
Therefore the remainder when 784{7^{84}} is divisible by 24022402 is −1 - 1 or 2402−12402 - 1 i.e., 24012401 .
The option (C) is correct.

Note: We know the property of power of a function a100{a^{100}} , we can write this as (a10)10{({a^{10}})^{10}} . Using this property we can simplify any harder power of a function. A combination is a selection of items from a collection, such that the order of selection does not matter. The formula of the combination nCr=n!r!(n−r)!^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}} . We use binomial theorems to solve harder problems and expand a function.