omerrauf wrote:

The answer should be C i.e. 2.

Here is how.

The expression is : \(3^{7^{11}}\)

Lets first resolve \(7^{11}\)

So \(7^2=49\)

So \(7^4=.....1\) where 1 is the last digit of the number

So \(7^8=.....1\) where 1 is the last digit of the number

So \(7^3=.....3\) where 3 is the last digit of the number

So \(7^{11}=7^3*7^8=........1*3=.............3\) where 3 is the last digit of the number

Now on to \(3^{7^{11}}=3^{......3}\) where 3 is the last digit of the exponent

We know the power is odd and its 3. Lets check out the last digit of some of the odd exponents for \(3\)

So \(3^1=3\)

So \(3^3=27\)

& \(3^5=....3\)

& \(3^7=.....7\)

& \(3^{11}=.........7\)

Recognize the patterns. It could be 3 or 7. Also, it is is 3 only in the case when the exponent is \(1\) or a multiple of \(5\)

We know the power is greater than \(1\) and not a multiple of \(5\) so the only possibility for the last digit is \(7\)

Now we know that \(\frac{7}{5}\) remainder is \(2\) Hence the answer must be C.

Well the answer posted above is correct, but the reasoning provided above is a little bit flawed.

If you look at \(3^{13}\) it has '3' as the units digit in exponent and if you try to solve it the answer will have

3 as the units digit, which when divided by 5 will give 3 as the remainder.

Here's another approach which I believe you can useUnderstand that in this question all you need to find out is the units digit of the expression \(3^{7^{11}}\). In order to do so, you must reduce this term in the form of \(3^x\).

\(3^x\) has a cyclicity of 4, i.e. the units digit of \(3^x\) repeats itself after four terms

\(3^1=

3\) --- so --- \(3^{4n+1}=...3\)------------------equation (1)

\(3^2=9\) --- so --- \(3^{4n+2}=...9\)------------------equation (2)

\(3^3=27\) --- so --- \(3^{4n+3}=...7\)------------------equation (3)

\(3^4=81\) --- so --- \(3^{4n+4}=...1\)------------------equation (4)

\(3^5=24

3\) --- so --- \(3^5=3^{4+1}=3^{4n+1}\)

So we can write any power of 3 in the form of \(3^{4n+k}\). This way calculating the value of 'k', we can easily find the units digit.

Here also we just need to write the power of 3 in 4n+k form.

Lets concentrate on \(7^{11}\)

If we divide this by 4, whatever we get will be the value of 'k' and our problem would be solved. Rewriting it as \((8-1)^{11}\)

Divide \((8-1)^{11}\) by 4 to get the value of k.

Here 8 will give the remainder 0 when divided by 4 and the only remainder we will get is from -1. Using the concept of negative remainders(which I'm assuming you know, incase you don't, feel free to ping me and I'll be happy to tell you) we'll get 3 as our final remainderHence \(7^{11}\) can be written as 4n+3.

So we can write our given expression \(3^{7^{11}}\) as \(3^{4n+3}\).

Using equation (3) above, we can easily make out that our units digit will be 7. Dividing this by 5 will give

2 as the remainder.

PS: Forgive me for my poor formatting. I'm still learning

Hope this helps.