Towering 7's !

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

What is the units digit of this expression?

77777777^{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
5 years, 9 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.

Answer is 3

Ashwani Karoriwal - 5 years, 9 months ago

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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.

So answer will be 3.

Aakash Kumar - 5 years, 9 months ago

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No, this doesn't work either. If k2(mod4)k \equiv 2 \pmod 4, what does that mean about its units digit? That doesn't mean it's 22, that actually means that it can be either 0,2,4,6,80, 2, 4, 6, 8. Is anybody checking these solutions before just upvoting them?

Michael Tong - 5 years, 9 months ago

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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 737^3, which has units digit 3. The solution is entirely correct, even if it leaves unstated that 7337^3 \equiv 3 at the end.

Caleb Stanford - 5 years, 9 months ago

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

Aakash Kumar - 5 years, 9 months ago

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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.

Peter Taylor Staff - 5 years, 9 months ago

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Thanks so much!!

Mark Mottian - 5 years, 9 months ago

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The unit's digit of the powers of 7 are 7,9,3,1,7,9,3,17,9,3,1,7,9,3,1\cdots. Therefore,7,9,3,17,9,3,1 continues like this.Now,we just want to calculate the exponent of 7 and we can figure out the answer.The exponent is

777777=?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 77 or 33. Now,note:

(i)If the exponent is of the form 4k+34k+3,the unit's digit is 3.

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

I am stuck now.

Rahul Saha - 5 years, 9 months ago

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

Mark Mottian - 5 years, 9 months ago

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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.

Will Wombell - 5 years, 9 months ago

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ANSWER IS 3

Shivam Rastogi - 5 years, 9 months ago

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Answer is 7

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

We can easily calculate 777^{7}=823543. So the units digit is 3.

Now,

(77)7(7^{7})^{7}=777^{7}x777^{7}x777^{7}..... {7 times}

We know that the units digit for 777^{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 (77)7(7^{7})^{7} --> 7 --- 1\boxed{1}

((By this I just mean to say that units digit of (77)7(7^{7})^{7} is 7))

Similarly,

((77)7)7((7^{7})^{7})^{7}=(77)7(7^{7})^{7}x(77)7(7^{7})^{7}x(77)7(7^{7})^{7}..... {7 times}

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

So, ((77)7)7((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--

777^{7} --> 3

(77)7(7^{7})^{7} --> 7

((77)7)7((7^{7})^{7})^{7} --> 3

(((77)7)7)7(((7^{7})^{7})^{7})^{7} --> 7

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

Upendra Singh - 5 years, 9 months ago

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Someone please post if there is a more efficient way of solving this problem. Anyways: Consider first 777^7. After testing out the first few powers, we see that the units digit has a cycle of four (7,9,3,17, 9, 3, 1). Since 73(mod4)7 \equiv 3 \pmod{4}, the last digit is 33. We continue for 373^7. The powers of 33 also have a cycle of four (3,9,7,13, 9, 7, 1). By the same logic, the last digit is 77. We now see a pattern of 7,3,7,3,7, 3, 7, 3, \ldots. Since 77 is raised 66 times, we must find the last digit for all numbers with 77 being raised an even number of times. We find that this number is 3\boxed{3}.

Hahn Lheem - 5 years, 9 months ago

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(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 7773(mod10){{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 (((77)7)7)7(((7^7)^7)^7)^7 \dots, but that is not how it is.

For example, 77=8235437^7 = 823543. The way you interpret it, 777=8235437{7^7}^7 = 823543^7, but actually it equals 78235437^{823543}.

Michael Tong - 5 years, 9 months ago

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

Maharnab Mitra - 5 years, 9 months ago

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@Maharnab Mitra 7odd(1)odd13(mod4)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 77's units digit cycles with a period of 4.

777733(mod10){{7^7}^7}^{\dots} \equiv 7^3 \equiv 3 \pmod {10}.

Michael Tong - 5 years, 9 months ago

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@Michael Tong Hi Michael. I'm quite new to modular arithmetic. Can you please explain how 777...7^{7^{7^{...}}} is congruent to 737^{3}?

Mark Mottian - 5 years, 9 months ago

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@Mark Mottian (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 77 have a cycle of four: 7,9,3,17, 9, 3, 1. Therefore, we will be using mod4\bmod{4}. Notice that: 7odd(mod4)(81)odd(mod4)(01)odd(mod4)(1)odd(mod4)1(mod4)1+4(mod4)3(mod4)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 777...7^{7^{7^{...}}} is obviously odd, so the value is congruent to 3(mod10)3 \pmod{10}.

Hope that helped.

EDIT: Oh and to answer your original question, we know that 77=73(mod4)73(mod4)7^7=7^{3 \pmod{4}} \equiv 7^3 \pmod{4} (remember that we are using mod4\bmod{4} because it has a cycle of four). 73(mod4)343(mod4)3(mod4)7^3 \pmod{4} \equiv 343 \pmod{4} \equiv 3\pmod{4}. Now we have 77...37^{7^{...^{3}}}, with an infinite number of 77s between the first 77 and the 33. We can continue this 737^3 thing infinitely, but the result will always stay as 737^3.

Hahn Lheem - 5 years, 9 months ago

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

Mark Mottian - 5 years, 9 months ago

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3

Raj Gupta - 5 years, 9 months ago

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Answer is 3.Michael tong has already explained.

Viswanath Reddy - 5 years, 9 months ago

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3

Vasu Subramanian - 5 years, 9 months ago

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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 !

Atreyee Chattopadhyay - 5 years, 9 months ago

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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

Rishabh Shukla - 5 years, 9 months ago

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

Prasad Nikam - 5 years, 9 months ago

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Correct.

Finn Hulse - 5 years, 9 months ago

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