# Towering 7's !

Can somebody suggest an easy solution (preferably without modular arithmetic) to this problem:

What is the units digit of this expression?

$$7^{7^{7^{7^{7^{7^{7}}}}}}$$

That is, 7 raised as a power six times (I'm not skilled in using Latex).

Thanks!

Note by Mark Mottian
4 years, 8 months ago

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An approach i dont find mentioned here:

As we know unit digit of power of 7 repeat in a cycle of 4 as (7,9,3,1,7,9,3,1...)

So we have to calculate the remainder of the power(six times 7) by 4.

7^(7^(^...))).. mod 4

=(-1)^(7^(^(....))) mod 4

i.e. -1 to the power an odd number, that's obviously -1

so the eqn becomes:

=(-1) mod 4

=3 mod 4

So the unit digit of the original question reduces to 7^(4k+3)

Using the cyclic nature of the powers of 7 (as mentioned at top) Unit digit is 3.

- 4 years, 8 months ago

7^1 = 7 , 7^2 =49 , 7^3 = 343 , 7^4 = 2401

so cycle length for units digit for power of 7 is 4.i.e (7,9,3,1)

if we can find remainder of 7(power raised five times) with 4 , units digit can be find easily.

7^(odd number) mod 4 = -1^(odd number) = -1 ,so remainder = 4-1=3

So answer will be 3rd element in list.

- 4 years, 8 months ago

No, this doesn't work either. If $$k \equiv 2 \pmod 4$$, what does that mean about its units digit? That doesn't mean it's $$2$$, that actually means that it can be either $$0, 2, 4, 6, 8$$. Is anybody checking these solutions before just upvoting them?

- 4 years, 8 months ago

I think you misunderstood the solution. The solution is to note that the exponent is congruent to 3 (mod 4), thus the expression itself is congruent to $$7^3$$, which has units digit 3. The solution is entirely correct, even if it leaves unstated that $$7^3 \equiv 3$$ at the end.

- 4 years, 8 months ago

remainder =3 So , answer is 3rd element in list which is 3.

- 4 years, 8 months ago

The unit's digit of the powers of 7 are $$7,9,3,1,7,9,3,1\cdots$$. Therefore,$$7,9,3,1$$ continues like this.Now,we just want to calculate the exponent of 7 and we can figure out the answer.The exponent is

$$7^{7^{7^{7^{7^{7}}}}}=?$$

Of course,that is the last thing we want to calculate.But note that this exponent is odd.Therefore,its unit's digit is either $$7$$ or $$3$$. Now,note:

(i)If the exponent is of the form $$4k+3$$,the unit's digit is 3.

(ii)If not,then the answer is 7.

I am stuck now.

- 4 years, 8 months ago

There's something about your solution that makes it easy to understand - especially if you're not familiar with modular arithmetic. Thanks!

- 4 years, 8 months ago

Hey Mark I edited the note for you with Latex. If you click the edit button you can see how I did it for yourself. You were pretty close, you just need to bracket the math with ( and you would have been good to go. Be sure to check out the formatting guide if you ever get stuck.

Staff - 4 years, 8 months ago

Thanks so much!!

- 4 years, 8 months ago

The answer is 3.Because of the repeating pattern.(7,9,3,1,....)

- 4 years, 8 months ago

Correct.

- 4 years, 8 months ago

let us calculate the last digit of 7^7; 7^7=(7^4)*(7^3)
as last digit of 7^4=1 and last digit of 7^3=3

so last digit of 7^7=3

and the loop continues at last we get 7^3 whose last digit is 3

- 4 years, 8 months ago

The last digit of 7^7 is 3,now raise the next seven to the power of 3,then raise the next seven to the power of the last digit of the last number found,continue this process and ultimately you'll get 3 as the answer !

- 4 years, 8 months ago

3

- 4 years, 8 months ago

- 4 years, 8 months ago

3

- 4 years, 8 months ago

Someone please post if there is a more efficient way of solving this problem. Anyways: Consider first $$7^7$$. After testing out the first few powers, we see that the units digit has a cycle of four ($$7, 9, 3, 1$$). Since $$7 \equiv 3 \pmod{4}$$, the last digit is $$3$$. We continue for $$3^7$$. The powers of $$3$$ also have a cycle of four ($$3, 9, 7, 1$$). By the same logic, the last digit is $$7$$. We now see a pattern of $$7, 3, 7, 3, \ldots$$. Since $$7$$ is raised $$6$$ times, we must find the last digit for all numbers with $$7$$ being raised an even number of times. We find that this number is $$\boxed{3}$$.

- 4 years, 8 months ago

(This is in mod 10)

7 = 7

7^7 = 3

7^7^7 = 7

7^7^7^7 = 3

7^7^7^7^7 = 7

7^7^7^7^7^7 = 3

7^7^7^7^7^7^7 = 7 (which is given in the problem) -- even though he says it's raised 6 times, he means it's raised 6 times over the original 7. According to you, that means the answer would be 7.

The correct answer lies in the fact that $${{7^7}^7}^{\dots} \equiv 3 \pmod {10}$$ no matter how many 7's you put on top of it. (As long as there is at least one)

You are interpreting the question as $$(((7^7)^7)^7)^7 \dots$$, but that is not how it is.

For example, $$7^7 = 823543$$. The way you interpret it, $${7^7}^7 = 823543^7$$, but actually it equals $$7^{823543}$$.

- 4 years, 8 months ago

Hey Michael, can you please show how you get the conclusion " no matter how many 7's you put on top of it"?

- 4 years, 8 months ago

$$7^{odd} \equiv (-1)^{odd} \equiv -1 \equiv 3 \pmod 4$$.

We look mod 4 not because it tells us about the units digit, but because $$7$$'s units digit cycles with a period of 4.

$${{7^7}^7}^{\dots} \equiv 7^3 \equiv 3 \pmod {10}$$.

- 4 years, 8 months ago

Hi Michael. I'm quite new to modular arithmetic. Can you please explain how $$7^{7^{7^{...}}}$$ is congruent to $$7^{3}$$?

- 4 years, 8 months ago

(I'm not Michael, and I am much worse than him in math, but I will do my best in explaining with more detail what Michael posted above.) The last digit of the powers of $$7$$ have a cycle of four: $$7, 9, 3, 1$$. Therefore, we will be using $$\bmod{4}$$. Notice that: $7^{\text{odd}} \pmod{4} \equiv (8-1)^{\text{odd}} \pmod{4} \equiv (0-1)^{\text{odd}} \pmod{4} \equiv (-1)^{\text{odd}} \pmod{4} \equiv -1 \pmod{4} \equiv -1+4 \pmod{4} \equiv 3 \pmod{4}$ The exponent of $$7^{7^{7^{...}}}$$ is obviously odd, so the value is congruent to $$3 \pmod{10}$$.

Hope that helped.

EDIT: Oh and to answer your original question, we know that $$7^7=7^{3 \pmod{4}} \equiv 7^3 \pmod{4}$$ (remember that we are using $$\bmod{4}$$ because it has a cycle of four). $$7^3 \pmod{4} \equiv 343 \pmod{4} \equiv 3\pmod{4}$$. Now we have $$7^{7^{...^{3}}}$$, with an infinite number of $$7$$s between the first $$7$$ and the $$3$$. We can continue this $$7^3$$ thing infinitely, but the result will always stay as $$7^3$$.

- 4 years, 8 months ago

That is very clear explanation for someone (like me) that is new to modular arithmetic. Thanks you for your rigorous explanation!

- 4 years, 8 months ago

I could do it by some logic and observation...

We can easily calculate $$7^{7}$$=823543. So the units digit is 3.

Now,

$$(7^{7})^{7}$$=$$7^{7}$$x$$7^{7}$$x$$7^{7}$$..... {7 times}

We know that the units digit for $$7^{7}$$ is 3. So talking just about the units digit, above expression is equivalent to

(....3)x(....3)x(....3)..... {7 times}

and units digit for this expression is 7.

((By (....3) I mean that the dots represent the previous digits(just for simplicity)))

In short $$(7^{7})^{7}$$ --> 7 --- $$\boxed{1}$$

((By this I just mean to say that units digit of $$(7^{7})^{7}$$ is 7))

Similarly,

$$((7^{7})^{7})^{7}$$=$$(7^{7})^{7}$$x$$(7^{7})^{7}$$x$$(7^{7})^{7}$$..... {7 times}

=(....7)x(....7)x(....7)..... {7 times} (from $$\boxed{1}$$)

So, $$((7^{7})^{7})^{7}$$ --> 3

You can go on like this and easily find out the answer. Also we can make out a pattern here to cut short the steps. The pattern is--

$$7^{7}$$ --> 3

$$(7^{7})^{7}$$ --> 7

$$((7^{7})^{7})^{7}$$ --> 3

$$(((7^{7})^{7})^{7})^{7}$$ --> 7

and so on we have-- 3 7 3 7 3 $$\boxed{7}$$ ANS

- 4 years, 8 months ago

- 4 years, 8 months ago

the easiest way would be to find the different unit ending for powers of seven and those are 1,3,7 and 9. then you could find the number of times 7 is multiplied, divide by four but keep the remainder, if the remainder is 1 then the end unit is 7 and so on.

- 4 years, 8 months ago