The odd digits in the completed grid were highlighted, producing the form of one of the creatures from the 1978 arcade game Space Invaders.

This is one possible solution path.
The last digits of T and W form *w*,
which is a two-digit fourth power, ie one of {16, 81},
so X = T + W ends in 7 or 9,
and as it’s a palindrome, X is one of {77, 99}.
But *x* is a square and therefore can’t end in 7,
so **X = 99** and ** w = 81**.
The only square that fits

*s* = 9S is now _49 in the grid,
and to be a multiple of 9 its digits must add up to a multiple of 9,
so it must be ** s = 549**,
making

We have T = _8 and W = _1, and they add up to 99,
so their first digits must add up to 9.
*q* has increasing digits,
so it must match one of {_18, _27, _36, _45},
starting with one of {1, 2, 3, 4}.
But N is a prime, so it ends in 1 or 3,
and it has the same digit sum as Q, ie 10,
which allows only {91, 73};
91 = 7×13 isn’t prime,
so **N = 73**.
Now *q* is one of {336, 345}
with T as one of {38, 48};
T is divisible by its digit sum,
but 38 isn’t divisible by 3 + 8 = 11,
so **T = 48**,
making ** q = 345**
and

From the grid, *n* is __5 (an odd multiple of 5)
and it’s a multiple of *x* = 49,
so it’s one of {245, 735}.
But we know A = *n* − K,
where K is an anagram of *n*,
so *n* > K and therefore *n* can’t be 245;
that leaves ** n = 735**.

K is one of {357, 375, 537, 573},
which when subtracted from *n* give
{378, 360, 198, 162} respectively for A,
but we can rule out the second of those
because *c* can’t start with 0.
Now *c* is a three-digit multiple of 49 starting with 2 or 8,
ie one of {245, 294, 833, 882},
for which the respective digit products are {40, 72, 72, 128}.
But because *z* is a two-digit multiple of *c*’s digit product,
*c*’s digit product must be < 50,
so ** c = 245**,

We know *d* = 2×55×735 + *a* ends in 7,
so *a* ends in 7, and because its digit product is 0
it must be ** a = 107**,
which makes

M = 2*f* has two digits,
so *f* < 50;
the possible triangular numbers are {10, 15, 21, 28, 36, 45}.
C is a palindrome (matching _404_) and can’t start/end with 0,
so 10 is ruled out;
*b* has the same last digit as *f*
and can’t be the prime 61, which rules out 21.
C is a multiple of *f* and the four possibilities for C/*f* are
{54045/15 = 3603,
84048/28 ≈ 3001.7,
64046/36 ≈ 1779.1,
54045/45 = 1201},
so **C = 54045**,
** b = 65** and

For V we have _8, which is the product of
its own two digits plus 18 (the digit sum of C).
That means V’s digit product ends in 0,
which is only satisfied by **V = 58**
(and 5×8 + 18 = 58, as expected).
Now *t* = 55×16 + 58,
ie ** t = 938**.
Then we have _9 for R, which is a factor of 58,
so

For L we have _37, so its digit sum is in the range 11 to 19.
The only Fibonacci number in that range is 13,
so **L = 337**.
For *i* we have __3 and it’s 735 plus a Fibonacci number,
so the Fibonacci number ends in 8 and is ≤ 993 − 735 = 258.
The only matching one is 8,
so ** i = 743**.

E = 55*u* − 5×649 is an obvious multiple of 5
and the first digit of *k* can’t be 0,
so E ends in 5, *u* is even,
and *k*, having a repeated digit, is one of {557, 577}.
J is 16 more than a cube, for which the three-digit options are
{141, 232, 359, 528, 745};
but its middle digit has to be 5 or 7,
so **J = 359** and ** k = 557**.

A = 162 and *j* have the same digit product,
ie 12, so *j* is one of {26, 34, 43, 62}.
The even values of *u* from 60 to 68 give
{55, 165, 275, 385, 495} for E,
of which only the last three could fit with *j*,
so *u* is one of {64, 66, 68}.
F = *u* + 1, ie one of {65, 67, 69},
which gives _69 for the palindrome *h*,
so ** h = 969**.
P =

*m* is a multiple of *a* = 107 with 3 as the middle digit,
for which the only possibility is ** m = 535**,
making

We know that *o* = 72 is the digit product of *r*,
which starts with 9, so its other two digits must have a product of 8;
since its digits are in descending order,
*r* is one of {942, 981}.
As *t* = 938 is the mean of *p* and *r*,
*p* and *r* are the same distance from 938,
in opposite directions;
if *r* is 981 = 938 + 43,
then *p* = 938 − 43 = 895,
but we know from the grid that *p* starts with 9.
Therefore ** r = 942**
and

The grid is now complete apart from H = 7__,
the two-digit * entry, and *g* = 9__.
The only values *g* could have, being 1 more than another entry,
are {935, 939, 943, 954, 970}.
From G’s clue we know that *g*’s digit product is a multiple of 54,
so the product of the second and third digits must be a multiple of 6,
which is only satisfied by ** g = 943**.

H matches 7__ and the sum of the digits entered so far is 364,
so H = 728 + 2×(the sum of the 3 unknown digits),
and from *v*’s clue we know it’s a multiple of 16.
The * entry is less than *u* = 68,
so H must end with one of {2, 4, 6},
the possibilities being {736, 752, 784}.
(In fact, if we were to look ahead at decoding the grid,
the second pair of digits is 28,
which can yield a letter from 1 to 26 only if * is less than 30,
forcing H to be 752.)
If H is 736 then the missing second digit of * would have to be
736/2 − (364 + 3 + 6) = −5, not possible;
if H is 784 then the missing digit would have to be
784/2 − (364 + 8 + 4) = 16, too big;
therefore **H = 752** and the missing digit is
752/2 − (364 + 5 + 2) = 5,
making *** = 25**.
This completes the grid.

Decoding the digit pairs in the grid modulo 25 (taking two rows at a time)
gives PCN__ODD__D __DIGITS__EVB __HIGHLIGHTED__ K__NISHIKADO__H.
With the odd digits highlighted,
we may recognise the form of one of the creatures from
the arcade game Space Invaders.
A little research will reveal that it was designed by
Tomohiro **Nishikado**.