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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.

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  • 2
    $\begingroup$ 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! :) $\endgroup$ – PiIsNot3 May 16 at 3:41
  • $\begingroup$ You may add the definition of pan-digital number (or put a link will do), I just know that term today $\endgroup$ – athin May 16 at 3:42
  • $\begingroup$ Thx..will do in the future..didn’t have time to learn it fully $\endgroup$ – Uvc May 16 at 3:43
  • $\begingroup$ Has a correct answer been given? If so, please don't forget to $\color{green}{\checkmark \small\text{Accept}}$ it :) $\endgroup$ – Rubio 2 days ago
  • $\begingroup$ 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. $\endgroup$ – Uvc 2 days ago
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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!

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  • $\begingroup$ ...yes......... $\endgroup$ – Uvc May 16 at 4:08
  • $\begingroup$ beautiful question!!! @Uvc +1ed $\endgroup$ – Omega Krypton May 16 at 4:09
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    $\begingroup$ 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). $\endgroup$ – Aranlyde May 16 at 4:25
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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, $$ \begin{gather*} (X - 1)^5 + (X + 7)^5 + (2X + 6)^5 + (4X + 3)^5 + (5X + 8)^5 \equiv 3 \! \! \! \pmod{10} \\ (X - 1) + (X + 7) + (2X + 6) + (4X + 3) + (5X + 8) \equiv 3 \! \! \! \pmod{10} \\ 13X + 23 \equiv 3 \! \! \! \pmod{10} \\ 3X \equiv 0 \! \! \! \pmod{10} \\ X \equiv 0 \! \! \! \pmod{10} \end{gather*} $$

Final answer:

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 $

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  • $\begingroup$ sorry, ninja-ed you, have an upvote! $\endgroup$ – Omega Krypton May 16 at 4:11
  • $\begingroup$ @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 :) $\endgroup$ – PiIsNot3 May 16 at 4:16
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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

$816725943$.

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  • $\begingroup$ sorry, ninja-ed you, have an upvote! $\endgroup$ – Omega Krypton May 16 at 4:11

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