In the completed grid, the 15 cells with an edge on the perimeter contained a ring of numbers in which the sum of each adjacent pair was a triangular number. Their letter equivalents spelt TRIANGLE SUMZ JOY, clockwise from the top.

This is one possible solution path. We’ll use “ac” to refer to across entries, “dl” for entries running down to the left, and “dr” for down-right entries.

Entry 2dr occupies exactly the same cells as 2ac,
plus one more cell (containing a single digit,
as it’s not one of the 15 perimeter cells).
From their clues GOT and TO we know 2dr = 2ac×G,
which is possible only if **G = 10** with 2dr ending in 0.
Now 6ac, 3dl and 1dr all have a factor of G^{2} = 100,
so they all end in 00, which we can enter in the grid.

14ac occupies four cells, the last of which is on the perimeter,
so it has four or five digits.
N^{(AN)} = (N^{N})^{A} is at least N^{N} (if A is 1)
and 7^{7} = 823543 is too big to fit, so N ≤ 6.
The minimum for N is 2,
because (1^{1})^{A} = 1, whatever A is.
The last two cells (two or three digits) of 14ac
are shared with 9dr = N^{3}×A^{2},
and 9dr is one digit shorter than 14ac.
If N is 2, then (2^{2})^{9} = 262144 is too big for 14ac,
so A ≤ 8 and 9dr ≤ 2^{3}×8^{2} = 512,
and 14ac would have to be a four-digit number;
the only possibilities for (14ac, 9dr) are
(1024, 200) if A is 5 and (4096, 288) if A is 6,
but the last two digits don’t match in either case.
If N is 3,
the only four- or five-digit value for (3^{3})^{A} is 19683,
but that would need A to be 3 too.
If N is 4,
the only value to fit 14ac is (4^{4})^{2} = 65536 with A being 2,
but that would give 4^{3}×2^{2} =
256 for 9dr where a four-digit number is needed.
If N is 5,
the only value to fit 14ac is (5^{5})^{1} = 3125,
with 5^{3}×1^{2} = 125 for 9dr: a match.
If N is 6,
the only value to fit 14ac is (6^{6})^{1} = 46656,
but that would give 6^{3}×1^{2} = 216
for 9dr where a four-digit number is needed.
Therefore **N = 5**, **A = 1**,
**14ac = 3125** and **9dr = 125**.

Now 9dl = ZO^{2} is 121_,
not ending in 0 because 17ac can’t start with 0.
Its prime factors must include (some of) {2, 3, 5, 7, 11, 13}
because there are no other prime factors in the numbers 1 to 15.
Factorising the possibilities gives
{1211 = 7×173,
1212 = 2^{2}×3×101,
1213 = 1213,
1214 = 2×607,
1215 = 3^{5}×5,
1216 = 2^{6}×19,
1217 = 1217,
1218 = 2×3×7×29,
1219 = 23×53},
which all have primes that are too large,
except for 1215 = 135×3^{2} = 15×9^{2}.
Z can’t be 135, so **Z = 15**,
**O = 9** and **9dl = 1215**.
That makes **5dr = 149**,
which must be entered with two digits in the first cell.

For 5dl in the grid we have 149_3,
and the clue tells us R and T are factors of it, so R and T are odd.
For 6ac = 900R we have __00 (four or five digits) in the grid,
so the third-to-last digit is the last digit of 9R, which is odd.
That’s also the first digit of 7dr = R + U + M + S = _0.
7dr can’t be 10 because that would need
{R, U, M, S} = {1, 2, 3, 4} in some order and 1 is already taken;
the maximum sum of four variables is 15 + 14 + 13 + 12 = 54,
so 7dr is either 30 or 50.
The possibilities for 6ac are now 6300 if R is 7 and 13500 if R is 15;
but 15 is already taken by Z, so **R = 7**,
**6ac = 6300** and **7dr = 30**.
The grid now gives **7dl = 36** = TY;
T is odd but can’t be 9 (taken by O),
so **T = 3** and **Y = 12**.

From 11dr, TERM = MARY, so 3E = 12, ie **E = 4**.
We have **2ac = 27** and **2dr = 270**,
making **3dl = 70000** = 17500S^{2},
so **S = 2**.
That makes **16ac = 24**.
Now 13dl = 7J ends in 2, so J ends in 6,
ie **J = 6**, **13dl = 42**
and **10dr = 157** (with two digits in the first cell).

Now 1dr = 12000, entered with 12 in the first cell.
For 8ac = 92 − M we have _9 in the grid, so M ends in 3,
ie **M = 13** (as 3 is taken by T),
which makes **8ac = 79**, **11dr = 1092**
and **1dl = 1209** (in three cells).
We know 7dr = 22 + U = 30, so **U = 8**,
confirming **11dl = 1157** and giving
**12dr = 7424** and **18dl = 511** (with 11 in its second cell).
For 12dl = 507I we have 709_ in the grid,
so **I = 14**, **12dl = 7098**.
The grid is now complete, apart from the corners,
and there’s only one letter value left,
so **L = 11**.

The numbers in the perimeter cells are
?, 7, 14, 1, 5, ?, 11, 4, 2, 8, ?, 15, 6, 9, 12.
The corresponding letters used in the clues are _RIAN_LE SU_Z JOY,
which could be completed with the three remaining letters as
**T**RIAN**G**LE SU**M**Z JOY;
hence the numbers 3, 10 and 13 are entered in the corners.
In the “roundabout” thus produced,
the sum of each pair of adjacent numbers in the perimeter cells
is a triangular number:
3 + 7 = 10, 7 + 14 = 21, 14 + 1 = 14, …,
9 + 12 = 21 and 12 + 3 = 15.