4

Finding remainder when $(2023)^{2023}$ is divided by $35$

My Try: We need to find remainder when $(2023)^{2023}$ is divided by $5$ and $7$

So here $2023=0\mod(7)\Longrightarrow (2023)^{2023}=0\mod(7)$

And $\displaystyle (2023)=3\mod(5)\Longrightarrow (2023)^{2023}=3^{2023}\mod(5)$

$\displaystyle (3)^2=(-1)\mod(5)\Longrightarrow (9)^{1011}=-1\mod(5)$

$\displaystyle 3(9)^{1011}=-3\mod(5)=2\mod(5)$

So remainder when $(2023)^{2023}$ is divided by $35$ is $0$

Which is obtained by multiplying remainder of a number $(2023)^{2023}$ divisible by $5$ and $7$. But answer is $7$

What's wrong with my solution,Please explain me, Thanks

Bill Dubuque
  • 272,048
jacky
  • 5,194

3 Answers3

2

You correctly deduced that

$$ (1) \qquad 2023^{2023} \equiv 2 \pmod 5, $$

and

$$ (2) \qquad 2023^{2023} \equiv 0 \pmod 7. $$

$$$$

However, your final step, "So remainder when $(2023)^{2023}$ is divided by $35$ is $0$ ", was not justified in your working and in fact is wrong. It is in general not true that

$\ a \equiv b\pmod m;\ a\equiv c \pmod n \implies a \equiv bc \pmod {mn}.$

$$$$ However, what we can say is this:

$(1)\implies 2023^{2023}\equiv x \pmod {35}\ $ where $\ x\in \{2,7,12,17,22,27,32\} $

and

$(2)\implies 2023^{2023}\equiv x \pmod {35}\ $ where $\ x\in \{0,7,14,21,28\}. $

So $x$ must be $7.$

Adam Rubinson
  • 20,052
1

Your mistake is in the final paragraph. $0\bmod7$ and $2\bmod5$ do not combine to $0\bmod35$. That would be $0\bmod7$ and $0\bmod5$. Instead, they combine to $7\bmod35$.

Note that $7\bmod35$ reduces modulo 7 to $0$ and it reduces modulo $5$ to $2$, which is what we're after. As for how to go the other way, you could use the Chinese remainder theorem, most proofs of it are constructive, and tells you exactly how to combine the moduli you have.

But you could also, in this case, note that we are after something divisible by $7$, and go through all five possibilities: $0,7,14,21,28$. Which one of them reduces to $2$ modulo $5$? That's likely always going to be the fastest method with a small modulus like $35$.

Arthur
  • 199,419
1

They pointed out your mistake. I think the systematic way is/was something like that:

You found $x=7k_1\tag 1$ and $x=5k_2+2\tag2$ Now since $\gcd(7,5)=1$ we should have $1$ as a linear combination: $3\times 5-7\times 2=1\tag3$ By using $(1), (2)$ and $(3)$ $$\begin{align} x&=15x-14x\\ &=15(7k_1)-14(5k_2+2)\\ &=35(3k_1-2k_2-1)+7 \end{align}$$ Hence, $x\equiv 7\pmod{35}.$

Bob Dobbs
  • 10,988
  • Please strive not to post more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Mar 29 '23 at 18:27
  • @BillDubuque Numbers are different in this question. Don't we repeat same questions in exams with different numbers? Then teachers are illegal. – Bob Dobbs Mar 29 '23 at 18:52
  • 1
    It doesn't matter - we already have all known methods of solving CRT (many hundreds of such).There is nothing new here. Further dupe answers to FAQs only serve to make it much more difficult for students searching to locate the "best" answers. – Bill Dubuque Mar 29 '23 at 18:59
  • @BillDubuque Okay Next time I pay attention to dupe questions. – Bob Dobbs Mar 29 '23 at 19:06