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What will be the units digit of $7777$ raised to the power of $8888$ ?

Can someone do the math with explaining the fact "units digit of $7777$ raised to the power of $8888$"?

Martin Sleziak
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mysterious
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  • Are you simply asking what the question means? All it's saying is, what's the last digit of $7777^{8888}$. For example, if the answer happened to be 1234567 (it's not), the answer is 7. – Michael Grant Apr 25 '13 at 15:04
  • This is not linear algebra, by the way. I've retagged it, which may also be a hint to some as to how to solve it. – Michael Grant Apr 25 '13 at 15:04

4 Answers4

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First, you only need the units digit of $7^{8888}$. The really simple thing is to look for the pattern: $7^0 \equiv 1, 7^1 \equiv 7, 7^2 \equiv 9, 7^3 \equiv 3, 7^4 \equiv 1 \pmod {10}$ and the pattern repeats. Since $8888$ is a multiple of $4$, $7777^{8888}$ ends in $1$.

To be a bit more formal, now we have $7777^{8888} \equiv 7^{8888} \equiv (7^4)^{2222}\equiv 1^{2222}=1 \pmod {10}$

Ross Millikan
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  • what is mod? does $X=y(mod\ c)$ mean $X=Kc+y$ where $K$ is any integer? – ABC Apr 25 '13 at 15:07
  • yes, that is what it means... – Michael Grant Apr 25 '13 at 15:09
  • @exploringnet: that is correct. Some insist on using $\equiv$ to mean equivalent, so I updated with that. The remainders on division by any $K$ form a ring, which means you can add, subtract, multiply (and maybe divide) like the integers. In some rings it is possible, as in $\pmod {10}$ to have $ab=0$ without $a=0$ or $b=0$-take $a=2, b=5$ – Ross Millikan Apr 25 '13 at 15:09
  • What is a ring.( sorry i'm a high school student) in the example you want to say $ab=0(mod\ 10)$? and can we operate these:$a=1(mod\ x)$ and $b=2(mod\ x)$ to get $$a[+,-,\times ,/]b=1[+,-,\times ,/]2 (mod\ x)$$ – ABC Apr 25 '13 at 15:14
  • @exploringnet: In abstract algebra we have groups, rings, and fields (and some more). Groups have one operation, are associative, and have inverses. Rings are like the integers in the way I said, but can be finite. You have two operations that act like addition and multiplication, additive inverses, and the distributive law. Fields are like the reals-you can add, subtract, multiply, and divide (except by 0). It forms a huge subject. – Ross Millikan Apr 25 '13 at 15:39
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Note that any integer $n$ can be written as $10a + b$ where $a$ and $b$ are integers and $b$ is between $0$ and $9$ (inclusive). Written in this way, $b$ is the units digit of $n$. For example $7777 = 10(777) + 7$; here $b = 7$.

Instead of raising a number to such a high power ($8888$), let's start with a smaller power, $2$. If we have written our number $n$ in the form $10a + b$, then we have $$n^2 = (10a + b)^2 = 100a^2 + 20ab + b^2.$$ But this is a positive integer and can be written in the aforementioned way. If $b^2 = 10c + d$ is it's representation, then $10(10a^2 + 2b + c) + d$ is the representation of $n^2$. Note that both $b^2$ and $n^2$ have the same units digit, $d$. We can summarise this by the following:

The last digit of $n^2$ is the same as the last digit of $(\text{the last digit of}\ n)^2$.

Using our number as an example, we can check this. Note that $$7777^2 = 60481729$$ which has last digit $9$. The last digit of $7777$ is $7$ and $7^2 = 49$, which also has last digit $9$.

Similar reasoning to the above (in particular the binomial theorem) can be used to prove the following generalisation, where $k$ is any positive integer:

The last digit of $n^k$ is the same as the last digit of $(\text{the last digit of}\ n)^k$.

So for the problem at hand, the above says that the last digit of $7777^{8888}$ is the same as the last digit of $7^{8888}$. The latter is still a huge number, but a little easier to deal with.

As I have explained in a previous answer, you can figure out what the last digit of $7^{8888}$ is in a fairly easy way. In the linked answer, the question is about the last digit of $2^{2006}$, but the process is the same, just with different numbers.

Community
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Michael Albanese
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  • Thnx I got you tottaly – mysterious Apr 25 '13 at 18:42
  • Let guess we have some numbers with power: $2^{5689}$, $3^{5689}$, $4^{5689}$, $5^{5689}$ ,$6^{5689}$ , $7^{5689}$ , $8^{5689}$, , $9^{5689}$ What I want to know is to evaluate the last digit after expanding the power. – mysterious Apr 25 '13 at 19:00
  • You can use the method I outline in the linked answer for all of these. Let me know if I have misunderstood your question. – Michael Albanese Apr 26 '13 at 03:32
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the units digit of a number is the same as the number mod 10

so we just need to compute $7777^{8888} \pmod {10}$

first $$7777^{8888} \equiv 7^{8888} \pmod {10}$$ and secondly by Eulers totient theorem $$7^{8888} \equiv 7^{0} \equiv 1 \pmod {10}$$ by Eulers totient theorem (since $\varphi(10)=4$ and $4 \mid 8888$)

so there's a quick way to see that the last digit is a 1

shobon
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$$7777^{1} \equiv 7 \pmod{10}$$

$$7777^{2} \equiv 9 \pmod{10}$$

$$7777^{3} \equiv 3 \pmod{10}$$

$$7777^{4} \equiv 1 \pmod{10}$$

$$7777^{5} \equiv 7 \pmod{10}$$

And the relation continues, in general you see that:

$$7777^{4n + 1} \equiv 7 \pmod{10}$$

And

$$7777^{4n} \equiv 1 \pmod{10}$$

$$\frac{8888}{4} = 2222$$

Hence, $7777^{8888} \equiv 1 \pmod{10}$.

Amad27
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