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Question: The remainder obtained \({{\left( 1! \right)}^{2}}+{{\left( 2! \right)}^{2}}+{{\left( 3! \right)}^{2...

The remainder obtained (1!)2+(2!)2+(3!)2+....+(100!)2{{\left( 1! \right)}^{2}}+{{\left( 2! \right)}^{2}}+{{\left( 3! \right)}^{2}}+....+{{\left( 100! \right)}^{2}} when is divided by 102{{10}^{2}} is:
A)27
B)28
C)17
D)14

Explanation

Solution

first we should know the values of 1!=1,2!=1×2=2,3!=1×2×3=61!=1,2!=1\times 2=2,3!=1\times 2\times 3=6….. and so on. And then divide each factorial term with 102. Whenever the factorial term is not divisible by 102 then we should consider the remainder of that particular sum of factorial square terms and that will be the resultant remainder.

Complete step-by-step answer :
From the problem we have to find out the remainder when (1!)2+(2!)2+(3!)2+........+(100!)2{{\left( 1! \right)}^{2}}+{{\left( 2! \right)}^{2}}+{{\left( 3! \right)}^{2}}+........+{{\left( 100! \right)}^{2}} is
divided by 100.
(1!)2+(2!)2+(3!)2+....+(100!)2{{\left( 1! \right)}^{2}}+{{\left( 2! \right)}^{2}}+{{\left( 3! \right)}^{2}}+....+{{\left( 100! \right)}^{2}}from this
⇒(1!)2=1×1=1{{\left( 1! \right)}^{2}}=1\times 1=1
⇒(2!)2=2×2=4{{\left( 2! \right)}^{2}}=2\times 2=4
⇒(3!)2=6×6=36{{\left( 3! \right)}^{2}}=6\times 6=36
⇒(4!)2=24×24=576{{\left( 4! \right)}^{2}}=24\times 24=576…and so on.
Here (4!)2=576{{\left( 4! \right)}^{2}}=576 is divisible by 100. But (1!)2=1,(2!)2=4,(3!)2=36{{\left( 1! \right)}^{2}}=1,{{\left( 2! \right)}^{2}}=4,{{\left( 3! \right)}^{2}}=36 are not divisible by
100.
So, from (4!)2{{\left( 4! \right)}^{2}} onwards all the factorials are divisible by 100.Which means for n>4n> 4 all divisible by 100.
It means that we have to find the remainder for the factorials of which n<4.n< 4.
Now, the given problem will reduce as shown below,
Remainder((1!)2+(2!)2+(3!)2+(4!)2100)\left( \dfrac{{{\left( 1! \right)}^{2}}+{{\left( 2! \right)}^{2}}+{{\left( 3! \right)}^{2}}+{{\left( 4! \right)}^{2}}}{100} \right)
Remainder(1+4+36+576100)\left( \dfrac{1+4+36+576}{100} \right)
Remainder(41+576100)\left( \dfrac{41+576}{100} \right)
Remainder(617100)\left( \dfrac{617}{100} \right)
Therefore,
The remainder obtained when (1!)2+(2!)2+(3!)2+....+(100!)2{{\left( 1! \right)}^{2}}+{{\left( 2! \right)}^{2}}+{{\left( 3! \right)}^{2}}+....+{{\left( 100! \right)}^{2}} is divided by 102{{10}^{2}} is “17”.
Hence the correct option is (C).

Note : In above solution we should consider the (4!)2{{\left( 4! \right)}^{2}} also to find the remainder.Without including this (4!)2{{\left( 4! \right)}^{2}}. Again we cannot find the remainder as the sum of those factorial square terms sum is below“100”. So that the absolute value of the remainder can be obtained.