## 777777^333333

Posted by Chris on June 25, 2013 – 10:54 am

What is the last digit of 777777^{333333}?

Although I’ve only used 7’s and 3’s, you should be able to generalise the 777777 and 333333 to get the same result.

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June 25th, 2013 at 1:52 pm

7. number of sevens doesn’t matter, number threes doesn’t matter as long as there are at least 2 threes. one 3, ends in a 3, two threes, ends in a 7, and everything on ends in a seven.

June 25th, 2013 at 2:16 pm

Hi bob. There’s an even more general answer, but what you’ve said is correct.

June 25th, 2013 at 9:42 pm

If I start with 777777^333333, I can rewrite this as (111111^333333)*(7^333333). Any integer ending in 1 raised to a positive integer power will have 1 in the last digit. Thus, the numbers of sevens and threes will have no effect on that part.

Seven raised to a power follows a modulus of 4 pattern:

7^0 = 1

7^1 = 7

7^2 = 49 (9)

7^3 = 343 (3)

7^4 = 2401 (1)

7^5 = 16807 (7)

…and so on. Thus the last digit will be a 1 when raised to a power divisible by 4. The last digit will be 7, 9, and 3, respectively, when raised to a power mod 4 of 1, 2, and 3.

As bob said, for at least two threes, the last digit of 77….77^33….33 is seven. In general for 77….77 raised to any positive integer power, the last digit will be 1, 7, 9, and 3 for the last two digits of the power modulus 4 of 0, 1, 2, and 3, respectively.

I hope that is general enough.

June 26th, 2013 at 7:04 am

Hi Chuck. Thanks for that analysis. I hadn’t thought of the split to 111111*7. I liked that.

I’ve been a bit mean about the general answer. Both you and bob gave the answer that I was after when I started writing that part of the problem. By the time I’d finished writing that bit, I knew that a more general answer was:

(10u +7)^(4v+1). Of course, there are many ways to get a number that ends in 7. i.e. why stick with a^b. So perhaps I was being silly.

I don’t know how bob did it. The way I did it was using modulo arithmetic. Then the string of 7’s may be replaced by a single 7 as 777777 = 7 (mod 10). For the power, somewhat excessively, I employ Euler’s theorem => a^φ(m) = 1 (mod m). φ(10) = 10(1 -1/2)(1 – 1/5) = 4. Then I verify that 4 (rather than 2, as 4 = 2*2) is the smallest power. As 7^2 = 9 = -1 (mod 10) => 4 is the smallest power. So 7^(4v+1) = 7 (mod 10). Now 333333 = 333300 + 32 + 1 = 1 (mod 4), hence 777777^333333 = 7 (mod 10) as does (10u+7)^(4v+1) (mod 10)