# Five Powers of Fives Produce Unique Pandigital Number…Solve for X..Tell me Y

Given: Y is a Pan-digital Number (no zero, 1 to 9 only) ending in 3.

Pan digital number consists of all 9 digits 1 to 9..each digit occurring only once as is the case here. Last digit is given as 3 and all other digits 1,2,4,5,6,7,8,9 can be anywhere in the number.

No googling, no computers, you don’t even need the calculator till the last step to calculate Y from X.

$$Y = (X-1) ^ 5 + ( X + 7 ) ^ 5 + ( 2X + 6 ) ^ 5 + ( 4X + 3 ) ^ 5 + ( 5 X + 8) ^ 5$$

The most concise and logical answer will be accepted.

• In the future, for these kinds of alphametic/mathematical puzzles, I strongly encourage you to use MathJax to make the formatting look good, as it does now. Here's a meta post from Mathematics SE that gives a quick tutorial. I also recommend reading the Markdown help page for other formatting matters. Keep up the good work! :) – HTM May 16 '19 at 3:41
• You may add the definition of pan-digital number (or put a link will do), I just know that term today – athin May 16 '19 at 3:42
• Thx..will do in the future..didn’t have time to learn it fully – Uvc May 16 '19 at 3:43
• Has a correct answer been given? If so, please don't forget to $\color{green}{\checkmark \small\text{Accept}}$ it :) – Rubio May 19 '19 at 11:20
• Explanations and answers given are very elaborate. What is required is concise logical answer. Once given, it will be accepted. I think it can be done with less than 100 characters easily. – Uvc May 19 '19 at 15:15

OK, now i realise its beauty...

for all $$i$$ from $$0$$ to $$9$$, $$i^5$$ mod $$10$$ = $$i$$

so to simply get the ending digit of $$X$$, the equation can be simplified:

$$Y = (X-1) + ( X + 7 ) + ( 2X + 6 ) + ( 4X + 3 ) + ( 5 X + 8)$$
$$Y=13X +23$$

More mod10-ing:

$$Y=3X+3$$

Sub $$Y mod10 = 3$$:

$$3=3X+3$$

Deducing $$X mod10$$:

$$3X=0$$
$$X=0 (mod 10)$$

Then, start from $$X =$$

$$0$$

Result (using a calculator in this very last step)

$$57593$$ Too small...

Next attempt: $$X =$$

$$10$$

Result (using a calculator in this very last step)

$$Y=816725493$$ Nice!

The result above is a pan-digital number as required by OP, so this is done!

• ...yes......... – Uvc May 16 '19 at 4:08
• beautiful question!!! @Uvc +1ed – Omega Krypton May 16 '19 at 4:09
• Technically, you don't even need trial and error, as rot13(trareny zntavghqr pnyphyngvbaf zrna vg pbhyq bayl or 10. 20 znxrf bar grez 108^5, juvpu vf zber gura 9 qvtvgf). – Aranlyde May 16 '19 at 4:25

Without computers or calculators (at least until the very end), the answer is

$$X = \boxed{10}$$

The key here is to realize that

the units digit of $$Y$$ being 3 limits our possibilities for $$X$$ by a lot. Finding the last digit of a positive integer is the same as taking the integer modulo 10, so we will take the given expression for $$Y$$ modulo 10, set it equal to 3, and solve for $$X.$$

To do this, we

apply Euler's Theorem, which states that for all coprime integers $$a, n$$ we have $$a^{\phi(n)} \equiv 1 \! \pmod{n},$$ where $$\phi(n)$$ is Euler's totient function. For this problem, we'll rely on a similar equation $$a^{\phi(n) + 1} \equiv a \! \pmod{n},$$ which works for any integers $$a, n,$$ not just coprime.

Applying this theorem:

We have $$n = 10,$$ so $$a^{\phi(10) + 1} \equiv a^5 \equiv a \! \! \! \pmod{10}.$$ Thus, $$\begin{gather*} (X - 1)^5 + (X + 7)^5 + (2X + 6)^5 + (4X + 3)^5 + (5X + 8)^5 \equiv 3 \! \! \! \pmod{10} \\ (X - 1) + (X + 7) + (2X + 6) + (4X + 3) + (5X + 8) \equiv 3 \! \! \! \pmod{10} \\ 13X + 23 \equiv 3 \! \! \! \pmod{10} \\ 3X \equiv 0 \! \! \! \pmod{10} \\ X \equiv 0 \! \! \! \pmod{10} \end{gather*}$$

We know now that $$X \equiv 0 \! \pmod{10}$$ i.e. $$X$$ is a multiple of 10 (0, 10, 20, 30, ...). Note that $$Y$$ has exactly 9 digits, so $$X = 0$$ can be ruled out since the sum will be less than 5 orders of magnitude. $$X = 10,$$ however, does have the potential to come close, and by using a calculator we find that indeed it does work. Any higher values of $$X$$ would cause it to have more than 9 digits, so this is our final and only answer.

For the record, the final solution for $$Y$$ is

$$816725493 = 9^5 + 17^5 + 26^5 + 43^5 + 58^5$$

• sorry, ninja-ed you, have an upvote! – Omega Krypton May 16 '19 at 4:11
• @OmegaKrypton Well you did get a five minute head start.. I was still typing out my MathJax when your answer popped up! No matters, the OP will decide who gets the check mark :) – HTM May 16 '19 at 4:16

So this puzzle hinges on the fact that

$$X^5mod10 = X$$. ($$1^5=1$$, $$2^5=32$$, $$3^5=243$$, and so on.)

This means that

$$(X+9)mod10+(X+7)mod10+(2X+6)mod10+(4X+3)mod10+(5X+8)mod10 = 3mod 10$$ (as $$(X-1)mod10 = (X+9)mod10$$).

This simplifies to

$$(13X+33)mod10=3$$. Because of how the 3-times table works, this only works if $$Xmod10=0$$.

Looking only at

number of digits (for order-of-magnitude calculation), $$8^5$$ (for $$X=0$$), doesn't work, but $$58^5$$ (for $$X=10$$) does, thus making it the only possible solution.

The number is therefore

$$816725943$$.

• sorry, ninja-ed you, have an upvote! – Omega Krypton May 16 '19 at 4:11