Cleopatra’s Needled by Ruslan
Puzzle explanation
The 2s in the initial grid formed the shape of a 4.
After replacing them all by 4s,
the sum of the grid entries was 4444444,
which appeared in in row 5 and in column 5.
Highlighting the remaining 4s produced a question mark;
so when Cleopatra got stuck on a clue, she would “question Mark” (Antony)!
Entries using the numerals that appear in Arabic writing
(٠ for 0,
١ for 1,
٢ for 2,
٣ for 3,
٤ for 4,
٥ for 5,
٦ for 6,
٧ for 7,
٨ for 8,
٩ for 9)
rather than the usual Western meaning of “Arabic numerals”
were accepted.
Entries indicating the question mark
by curves drawn through the relevant cells,
rather than shading, were also accepted.
Puzzle solution process
This is one possible logical path for solving the puzzle.
We’ll use the notation A → E for an answer and its grid entry.
For any answer with two digits, ab,
the answer 10a + b → the entry 11a + 2b.
A threedigit answer 100a + 10b + c → 101a + 11b + 2c.
For 9ac, the answer is one digit, let’s say a → 2a.
The entry has two digits, so a ≥ 5.
Obviously, 2a ends with an even digit;
a is that digit + 2, so it’s also even, ie, either 6 or 8.
2 × 6 = 12, 2 + 2 = 4, not 6, so that doesn’t work;
2 × 8 = 16, 6 + 2 = 8 does work, so 9ac = 8 → 16.
For 8ac, we have ab → ac, where c = b^{2},
so b is one of {0, 1, 2, 3} and 11a + 2b = 10a + b^{2}.
That gives a = b × (b − 2).
If b = 1, a = 1 (not a digit);
if b = 0 or 2, a = 0 (not allowed for a first digit);
thus b = 3 and a = 3, so 8ac = 33 → 39.
For 18ac, we have ab → ba, so 11a + 2b = 10b + a.
That gives 5a = 4b, which makes b = 5 and a = 4, so 18ac = 45 → 54.
For 17ac, 1 + 2 × (grid entry) has two digits,
so the grid entry 11a + 2b ≤ 49, which restricts a to the range [1, 4].
If a = 4, the answer 40 + b → the grid entry 44 + 2b ≤ 49, so b would be ≤ 2;
that restricts the clue answer to the range [10, 42].
Using e, f for the two digits whose product is the clue answer,
and c, d for the two digits of the grid entry,
we have 10e + f = 1 + 2(10c + d) = 10(2c) + (2d + 1).
If d < 5, then 2d + 1 gives a single digit,
so we can match f = 2d + 1 (odd) and e = 2c (even).
If d ≥ 5, then 2d + 1 gives a “carry” of 1,
making f = 2d + 1 − 10 (odd) and e = 2c + 1 (odd).
Both e and f are ≥ 2 (or no twodigit product is possible),
so f is one of {3, 5, 7, 9}.
The possible values for 10e + f that give a twodigit product e × f ≤ 42,
are {25, 27, 29, 35, 37, 39, 43, 45, 47, 49, 53, 55, 57, 63, 65, 67, 73, 75, 83, 85, 93}.
Of those, only 45 works, giving 17ac = 20 → 22.
These answers allow some others to be calculated immediately,
giving 10ac = 3187 → 3206,
14dn = 1188 → 1206
and 16dn = 236 → 247.
Knowing 10ac, we can now construct 20ac = 713183 → 713206
and 6dn = 378 → 396.
For 2dn, we know the answer is 4a5a, where a is the inserted/added digit.
Algebraically, that’s 4050 + 101a,
which → a grid entry of 4050 + 101a + 4 + 5 + 2a = 4059 + 103a.
From the grid, we know the third digit is 6,
so a is one of {1, 2, 3}, the grid entry is one of {4162, 4265, 4368}
and the clue answer is one of {4151, 4252, 4353}.
For 15dn, the answer is then one of {251, 252, 253}
(the 2 is from the first digit of 16dn, the rest from the end of 2dn);
from the grid we know the entry is _5_,
which rules out 252 → 261 and 253 → 263.
So 15dn = 251 → 259,
which makes 2dn = 4151 → 4162.
We can now calculate 5dn = 621 → 630.
For 15ac, the clue answer is a square and the grid entry is 22__,
so the square must start with 21 or 22.
46^{2} = 2116 → 2126, which is too small,
and 48^{2} = 2304 is too large,
so it must be 47^{2}:
15ac = 2209 → 2222.
We can now calculate 7ac = 1309 → 1322
and construct 11dn = 22208 → 22222.
That in turn allows construction of 3dn = 43216 → 43232.
For 12ac, we know the grid entry is 23_2_, so the clue answer must start with 2.
The first stage of constructing the answer is 3dn − 1 = 43215,
so the amount by which to reduce the first and third digits must be 2,
giving 12ac = 23015 → 23026.
For 1dn, the clue answer has 3 digits and the grid entry has 4,
so it must be 9__ → 1___.
That makes the grid entry for 1ac 144__6.
Its clue answer is 93 less than a square,
so the square root can’t be too far from √144006 ≈ 379.5.
Trying 379, we get 379^{2} − 93 = 143548 → 143573 (too small),
and 381^{2} − 93 = 145068 (too big),
so it must be 380^{2} − 93: 1ac = 144307 → 144326.
For 4dn, the grid entry is 32302_3, so the digit affixed at both ends must be 3,
and we can construct the answer as 4dn = 3230183 → 3230203.
For 13dn, the grid entry is 62_0 and the missing digit is even.
It can’t be 0 because an entry of 6200 needs a clue answer of 61__, but no odd digits are allowed.
For 19ac, the grid entry is one of {1022, 1024, 1026, 1028}.
The threedigit clue answer must be 9__ and, since 996 → 1020 is too small,
it’s one of {997, 998, 999}, with corresponding grid entries {1022, 1024, 1026}.
The corresponding fourdigit answers are {1015, 1016, 1017}
and the corresponding sums of the two answers are {2012, 2014, 2016}
(each a permutation of the corresponding grid entry).
Trying these in turn, 2012 isn’t divisible by the 5 from the answer 1015,
and 2014 isn’t divisible by the 6 from 1016 (since it’s not divisible by 3),
which only leaves 19ac = 999 → 1026
and 19ac = 1017 → 1026,
confirmed by 2016 = 9 × 224 and 2016 = 7 × 288.
The grid entry for 13dn is now 6260,
confirmed by a clue answer of 13dn = 6244 → 6260, with all even digits.
For 1dn, the fourdigit answer from 19ac gives 1017 − 8 = 1009, not a threedigit number,
so 1dn = 991 → 1010,
and the initial grid is complete.
The most frequent digit in the grid is 2: there are 16 of them, making the shape of a 4.
Replacing all the 2s with 4s, the sum of the entries in the revised grid is 4444444,
which appears in row 5 and column 5.
The other six 4s make an approximate questionmark shape,
so what Cleopatra had to do when “needled” by a clue
was to “question Mark” (ie, Mark Antony).
