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!

  • $\begingroup$ This is the most intellectually satisfying math puzzle I created to date, irrespective of the votes $\endgroup$ – Uvc Jun 28 '19 at 11:57
  • $\begingroup$ @Uvc You .. probably need to figure out how to delete your own comments. The result of others tidying up after themselves when you don't do so also is like listening to one side of a phone conversation and trying to figure out what's happening. Web, mobile-web, and app all provide you the ability to delete comments when their usefulness has ended; please delete them. $\endgroup$ – Rubio Jul 2 '19 at 0:35
  • $\begingroup$ Will look into.. $\endgroup$ – Uvc Jul 2 '19 at 1:46
  • $\begingroup$ Installed the app..now I think I can $\endgroup$ – Uvc Jul 2 '19 at 2:05




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.

  • $\begingroup$ Excellent!!...you have succeeded..I have accomplished my mission.. $\endgroup$ – Uvc Jun 28 '19 at 11:54

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:


  • $\begingroup$ 64+27=91 works too $\endgroup$ – JMP Jun 28 '19 at 10:48
  • $\begingroup$ oops, missed that, thanks! @JonMarkPerry $\endgroup$ – Omega Krypton Jun 28 '19 at 10:49
  • $\begingroup$ and 27+27=54, 8+8=16 $\endgroup$ – JMP Jun 28 '19 at 10:50
  • $\begingroup$ Will eliminate different possibilities..Sum of 2 Different cubes..will edit that $\endgroup$ – Uvc Jun 28 '19 at 10:52
  • $\begingroup$ 123<= third term<=178, so 91 gets eliminated! (as power must be 4) $\endgroup$ – JMP Jun 28 '19 at 10:58

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