Listen to my Story…Let us find the Unique Invisible Pan Digital Pair

Question

I was an avid reader of Popular Science magazine. In the last page or so, they usually had visual clues without words to make useful stuff.

I always wanted to create a mathematical puzzle like that with no expressions.

Let us see whether I succeeded. Sometimes, I do not wear puzzle solvers hat fully..I can supply the Missing info, only if needed.

The story goes like this:

We have two separate zero less Pan Digital Expressions with all digits from one to nine occurring exactly once on each side, excluding the power.

All the clues stated below apply to both expressions unless stated otherwise.

On the left side, you have four numbers raised to the same power and summing up. This power is a square of a digit.

On the right side you have Pandigital number which is different for the two expressions. However their first and last digits match..you don’t need this to solve..might be useful to verify.

Let us come back to left side and address each of the four terms.

First term contains a two digit cube.

Second term contains two digits and is sum of two Different cubes.

Third term contains a three digit number.

Fourth term contains a two digit number.

Two Pan Digital Expressions have same first, second, third terms. Fourth term contains reversed digits..that is the only difference on left side..leading to two different Pan digitals.

You can use the calculators few times to do the final figuring out of the Pan digital pair.

End of the story...Good Luck!

Answer 1

Solution:

$27^4+35^4+149^4+68^4=516297843$
$27^4+35^4+149^4+86^4=549617283$

Logic:

We know $HI^4-IH^4 \equiv 0 \pmod{10}$ as the last digit doesn't change. Over fourth powers we have $1\to1,2\to6,3\to1,4\to6,5\to5,6\to6,7\to1,8\to6,9\to1$, but we cannot have $1$ or $5$, so the only possibilities for $H$ and $I$ are $37,39,79, 24, 26, 28, 46, 48, 68$. Coupled with the limits for the first two terms, and the upper bound of $176$ for the third term, the search area is vastly reduced to about $20$ cases.

Answer 2

Clarification of the Problem and Very Partial Answer

Transforming the story into the expression. A through I are distinct digits from 1 to 9, X and Y different digits within 1 to 9, and m and n any integer.

AB^n^2 + CD^n^2 + EFG^n^2 + HI^n^2 = X???????Y and another set of
AB^m^2 + CD^m^2 + EFG^m^2 + IH^m^2 = X???????Y

AB can be

27 or 64 (2-digit cube)

CD can be

1+27=28, 1+64=65, 8+27=35, 8+64=72, or 27+64=91 (sum of 2 cubes)

m and n

2 $\leq$ m,n since otherwise the L.H.S's cannot reach a 9-digit number (thanks @JonMarkPerry)
m,n $\leq$ 9 as stated in question
Therefore 2 $\leq$ m,n $\leq$ 9

EFG can then only be

within the range 123 $\leq$ EFG $\leq$ 176 since otherwise EFG^(m or n)^2 would be 10+ digits long (thanks @JonMarkPerry).
EFG $\not\equiv$ 1 ($\mod{10}$) since E $=$ G

Exhaustive list of possible values of EFG:

123,124,125,126,127,128,129

132,134,135,136,137,138,139

142,143,145,146,147,148,149

152,153,154,156,157,158,159

162,163,164,165,167,168,169

172,173,174,175,176