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By Euler's theorem, $a^{\varphi(100)} \equiv\ 1 \pmod {100}$.

We know that the last two digits of $9^{40}$ are non-zero. So they can't even be $01$.

Since $1\equiv\ 1 \pmod {100}$, how come $9^{40}\equiv\ 1 \pmod {100}$?

I have looked at: Find the last two digits of $9^{{9}^{9}}$,

Find the last two digits of the number $9^{9^9}$ ,

Find the last two digits of $9^{9^{9}}$

Community
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AkaiShuichi
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  • "We know that the last two digits of $9^{40}$ are non-zero." That doesn't mean neither digit is zero; it just means they can't both be zero. So why do you think it can't be $01$? – Erick Wong Dec 22 '15 at 17:14
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    you're right i was thinking only of the the units digit – AkaiShuichi Dec 22 '15 at 17:17
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    @pyUser: When a more experienced user edits your post, learn from it, don't roll back to your own improperly formatted version. – Alex M. Dec 22 '15 at 17:18
  • @AlexM. can you teach me how to apply edits from one version to another one? or did you just manually MathJax the thing? – gt6989b Dec 22 '15 at 17:19
  • You quoted one way of doing it. We have $\varphi(100)=\varphi(25)\varphi(4)=40$. So $9^{40}\equiv 1\pmod{100}$, the last two digits, going righttwards, are $0$ and $1$. – André Nicolas Dec 22 '15 at 17:24
  • @gt6989b: I've done it manually, I don't know if it is possible to do it in a more sophisticated way. – Alex M. Dec 23 '15 at 11:17

7 Answers7

10

By binomial theorem:

$$(10-1)^{40} = \sum_{i=0}^{40} \binom{40}{i}(-1)^{40-i}10^{i}$$

Modulo $10^2$, you only need to look at $i=0,1$.

So: $$9^{40}\equiv (-1)^{40} + \binom{40}{1}(-1)^{39}\cdot 10 \equiv (-1)^{40}\equiv 1\pmod{100}$$

Note that this works for $9^{10}$, too.

Thomas Andrews
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5

$9^2\equiv\ -19\ (mod100)$

$9^4 \equiv\ 61\ (mod 100)$

$9^8\equiv\ 21 \ (mod 100)$

$9^{10}\equiv\ 1 \ (mod 100)$

$9^{40}\equiv\ 1 \ (mod 100)$

2

Calculate $$9^{10} = 3486784401$$ So $$9^{40} \mod 100 \equiv (9^{10}\mod 100)^4 \mod 100 $$ $$\equiv(1)^4 \mod 100 \equiv 1 \mod 100$$

amcalde
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It is straightforward $9^{40} = 81^{20} = 6561^{10} \equiv 61^{10} \pmod{100} = 3721^5 \pmod{100} \equiv\ 21^5 \pmod{100}$
$= 441 \times 441 \times 21 \pmod{100}\equiv\ 41 \times 41 \times 21 \pmod{100}$
$= 35301 \pmod{100} \equiv\ 1 \ \pmod{100}$

and this is just one of the many many routes you could take.

AnalysisStudent0414
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Note that $$99^2=9801\equiv 1\pmod{100}\implies99^{40}\equiv 1\pmod{100}\implies9^{40}11^{40}\equiv 1\pmod{100}$$

$$9^{40}\equiv 1\pmod{100}$$

AsdrubalBeltran
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If you want to avoid the equation

$9^{40}=147808829414345923316083210206383297601$,

the question is how low you prefer your numbers to be.

One alternative is using $40=4*2*5$ (and 7 digits for the last operation).

$9^4=61$ (mod 100)

$61^2=21$ (mod 100), and

$21^5=01$ (mod 100)

Mikael Jensen
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0

To answer your question, we know the last digit cannot be 0, as this would make it divisible by 10. There is no such restriction on the second to last digit.

When taking mod 100, one must divide the result by 100, so $9^{40}$ becomes much smaller. Euler's theorem says that this much smaller value is 1, and this can be verified by computation as shown in the other answers.

Abc Bcd
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