Prove that $53^{53}-33^3$ is divisible by $10$

I don't know modular arithmetic, so I tried things like that:

$53^3 \cdot 53^{50}-33^3=(33+20)^3 \cdot 53^{50}-33^3=(33+20)(33+20)(33+20)\cdot 53^{50}-33^3$ but I get stuck and don't know how to continue

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    $\begingroup$ One problem is that Prelude> (53^53 - 33^3) `mod` 10 ~> 6 it isn't divisible by $10$. $\endgroup$ – Daniel Fischer Nov 17 '13 at 12:23
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    $\begingroup$ Try to find out the last digit of $53^{53}$ and of $33^3$. If they are the same, your number is divisible by 10. To figure out the digits, do all calculations mod $10$! $\endgroup$ – Leif Sabellek Nov 17 '13 at 12:25
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    $\begingroup$ I am guessing he means $53^{53}-33^{33}$, which is divisible by $10$. $\endgroup$ – L. F. Nov 17 '13 at 12:25
  • $\begingroup$ Well, MR.BEAN does not know modular arithmetic. Any way the problem can be solved without modular arithmetic?? $\endgroup$ – K. Rmth Nov 17 '13 at 12:41

One way to hide the omnipresence of modular arithmetic is to note that a number is divisible by $10$ if and only if the final digit is $0$; and that the final digit of any number formed by a $\{+,-,\times\}$-recipe depends only on the final digits of the ingredients.

Now, the final digits of the powers of $3$ are $1$, $3$, $9$, $7$, and then the final digits repeat, starting with $81=3^4$. So the $4k$-th power of any number ending in $3$ will end in $1$, and thus $53^{53}=53^{52}\cdot53$ ends in $3$. On the other hand, $33^{33}=33^{32}\cdot33$ also ends in $3$, their difference ends in zero. If, as @MR.BEAN stated, the problem is $53^{53}-33^3$, then since $33^3$ ends in $7$, it’s the sum that is divisible by $10$.

  • $\begingroup$ You beat me to it by 2 minutes! $\endgroup$ – John Bentin Nov 17 '13 at 13:58
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    $\begingroup$ With about eight times as many words, as usual... $\endgroup$ – Lubin Nov 17 '13 at 14:00

The best way to prove this is to learn modular arithmetic. Any other strategy is simply a waste of time. So your first task is this:

Work out why $53^{33}$ is the same as $3^{33}$ mod 10.

Let us know when you understand that, and we can proceed to the next lesson.


$3^4=81,$ which ends in a $1$, and any power of a number that ends in $1$ also ends in $1$. Hence $3^{52}=(3^4)^{13}$ ends in $1,$ and so $53^{53}$ ends in $3$. Similarly, $33^{33}$ ends in $3$. So their difference ends in $0$.


Using Binomial Theorem, $\displaystyle(53)^{53}=(50+3)^{53}=3^{53}+\sum_{1\le r\le53}\binom{53}r50^r3^{53-r}$ $\displaystyle=3^{53}+10A$ as $\displaystyle\sum_{1\le r\le53}\binom{53}r50^r3^{53-r}$ is divisible by $10$

Now, $3^1=3,3^2=9,3^3=27,3^4=81$

Observe that the remainder by $10,$ follows a cycle $3,9,7,1$

So, we can conclude that $3^{53}\%10$ will be $3$ as $53\%4=1$

As $\displaystyle53\equiv3\pmod{10}\implies 53^{53}\equiv 3^{53}$

Similarly, $\displaystyle33^3=(30+3)^3=3^3+\sum_{1\le r\le3}\binom3r30^r\cdot3^{3-r}=3^3+10B$

as $\displaystyle\sum_{1\le r\le3}\binom3r30^r\cdot3^{3-r}$ is divisible by $10$

Now. $\displaystyle3^3\%10=7$

  • $\begingroup$ read my question well: I don't know modular arithmetic $\endgroup$ – MR.BEAN Nov 17 '13 at 14:35
  • $\begingroup$ @MR.BEAN, please find the edited version $\endgroup$ – lab bhattacharjee Nov 17 '13 at 16:37

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