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

Question

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.

Answer 1

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!

Answer 2

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 $

Answer 3

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

$816725943$.