Question

Let f(n) = 1 × 3 × 5 × · · · × (2n - 1). So, f(1) = 1; f(2) = 1 × 3; f(3) = 1 × 3 × 5 and so on.
Compute the remainder when f(1) + f(2) + f(3) + · · · + f(2016) is divided by 100.

Answer 1

Answer:

The remainder when f(1) + f(2) + f(3) + · · · + f(2016) is divided by 100 is 24

Step-by-step explanation:

Given that

$f(n)=1\times3\times5\times...........\times(2n-1)$

Now, $f(1)+f(2)+f(3)+......+f(2016)$

$=1+(1\times3)+(1\times3\times5)+(1\times3\times5\times7)+(1\times3\times5\times7\times9)+(1\times3\times5\times7\times9\times11)+(1\times3\times5\times7\times9\times11\times13)+(1\times3\times5\times7\times9\times11\times13\times15)+..........+(1\times3\times5\times..........\times4031)$

$=1+3+15+105+945+10395+135135+(81081\times25)+.....+\text{upto 2016 terms}$

If we divide 1, 3, 15, 105, 945, 10395,... by 4 we get the remainder as 1, 3, 3, 1 and this pattern repeats every four terms.

1+3 = 4 which is divisible by 4 and 2016 divided by 4 is 504 so sum of every two terms will be divisible by 4

Also, after the first 7 terms, all the terms are divisible by 25

So, if we take out the first 8 terms, the remaining sum of 2008 terms will both be divisible by 25 and 4, therefore divisible by 100.

Now we have to check the remainder when the sum of first 8 terms is divided by 100

We know that any number with the last two digits being 0 is divisible by 100

Adding the last two numbers of the first 8 terms, we get

1+3+15+05+45+95+35+25 = 224

And 224 = 200+24

∵ 200 is divisible by 100, in the end we get the remainder as 24.

I hope this answer helps.

Answer 2

Step-by-step explanation:

The remainder when f(1) + f(2) + f(3) + · · · + f(2016) is divided by 100 is 24

Step-by-step explanation:

Given that

f(n)=1\times3\times5\times...........\times(2n-1)f(n)=1×3×5×...........×(2n−1)

Now, f(1)+f(2)+f(3)+......+f(2016)f(1)+f(2)+f(3)+......+f(2016)

=1+(1\times3)+(1\times3\times5)+(1\times3\times5\times7)+(1\times3\times5\times7\times9)+(1\times3\times5\times7\times9\times11)+(1\times3\times5\times7\times9\times11\times13)+(1\times3\times5\times7\times9\times11\times13\times15)+..........+(1\times3\times5\times..........\times4031)=1+(1×3)+(1×3×5)+(1×3×5×7)+(1×3×5×7×9)+(1×3×5×7×9×11)+(1×3×5×7×9×11×13)+(1×3×5×7×9×11×13×15)+..........+(1×3×5×..........×4031)

=1+3+15+105+945+10395+135135+(81081\times25)+.....+\text{upto 2016 terms}=1+3+15+105+945+10395+135135+(81081×25)+.....+upto 2016 terms

If we divide 1, 3, 15, 105, 945, 10395,... by 4 we get the remainder as 1, 3, 3, 1 and this pattern repeats every four terms.

1+3 = 4 which is divisible by 4 and 2016 divided by 4 is 504 so sum of every two terms will be divisible by 4

Also, after the first 7 terms, all the terms are divisible by 25

So, if we take out the first 8 terms, the remaining sum of 2008 terms will both be divisible by 25 and 4, therefore divisible by 100.

Now we have to check the remainder when the sum of first 8 terms is divided by 100

We know that any number with the last two digits being 0 is divisible by 100

Adding the last two numbers of the first 8 terms, we get

1+3+15+05+45+95+35+25 = 224

And 224 = 200+24

∵ 200 is divisible by 100, in the end we get the remainder as 24.

I hope this answer helps.