Odd One Out by Oyler
Puzzle explanation
First of all, construct the third grid. The across bars are readily fixed. The central column must have no bars since the only unused locations would give an even number of entries. The six bars to be added must be 2 1 0 1 2 or 1 2 0 2 1. The only way to place two bars in the first column would block off a cell. Hence the doubles are in columns 2 and 4, where there is only one unused pair. In the first column, only one of the two choices gives seven entries.
Number the two given grids as I and II from the left, and the one just constructed as III.
A(II) cannot be prime from the d set of clues since 5G must refer to the unknown grid which leaves A/b in grid II. Taking A(II) to be Fibonacci gives 1597, 2584, 4181 and 6765 as possibilities. The b could be Fibonacci or 2E. Taking Fibonacci with Fibonacci yields no solutions as does 1597 with 1019, 2584 with 2029, 4181 with 4049 and 6765 with 6069 thus A(II) is the cube. Dividing possible cubes by relevant Fibonacci numbers again yields no solution thus b(II) is 2E and so even. We can therefore eliminate 1331, 2197, 3375, 4913, 6859 and 9261 leaving 1728 = 12³, 2744 = 14³, 4096 = 16³ and 5832 = 18³. 1728 can be discounted since if b = 12 then E = 6 and there are no leading zeros; b = 18 is also ruled out in that way. 2744 goes since the only appropriate factor leads to d < 100. 4096 also goes since continually multiplying by 2 does not give a number in the 40s. Therefore the cube, A(II) is 5832, b(II) is 54, d(II) is 108 and E(II) is 27.
Turn attention to the c clue d´–B. It cannot be grid I since c is 4 digit and both B and d are 2 digit. We can deduce that it cannot go in the unknown grid since B must have 3 digits in that grid and c must have 5. So d´–B is grid II and B(II) is 441 with c(II) 360.
D(II) is not the palindrome and C(II) must be the prime and is either 61 or 67.
If we assume G(II) is A/3 then we get 1944 and D(II) as EG does not fit so would have to be g(A–f) with g as 34 or 64, from the g clues. Testing the fs gives 6 digit entries thus G(II) is not A/3 so that must go with the A as a Fibonacci. We have deduced that A(III) is 2 digit and if it is Fibonacci then only 21 is divisible by 3 which gives 7 and single digit which is not possible. Thus A(I) is Fibonacci and divisible by 3. The two possibilities are 144 and 987. Only the latter gives a 3 digit entry when divided by 3 so A(I) is 987, G(I) is 329 and G(II) must be prime.
Now c(III) is 5 digit so b(III) can’t be; thus b(I) is AB. 70003/987 = 70.9… and 79993/987 = 81.0… and the required number must end in 9 so B(I) is 79, a prime and b(I) is 77973.
Now A+e´ must be F(III), F(I) cannot be 2B so F(II) must be and so is 882. Now g(II) is either 81 a square or 89 Fibonacci so a(II) is not G–A nor is a(I) so a(III) is. D(II) must be EG and the product of 27 with a prime. Let us assume that a(II) is G´ so G(II) is a prime of the form 2×41 or 2×49. Of the six possibilities only 2741 gives a D that fits 74007. However f(II) is neither prime, palindromic or twice a square so our assumption is incorrect. Therefore a(II) is a palindrome 2442 and D(II) starts with a 4. Testing primes yields 1667 and 1741 as possibilities. However only the latter fits with g(II) and f(II). So D(II) is 47007, g(II) is 81 a square and f(II) 777 a palindrome.
A(III) is a 2 digit prime and b(III) a 3 digit Fibonacci and G(III) is b/A and 2 digit. The only answer is that A(III) is 13, b(III) is 377 and G(III) is 29.
D(I) must be g(A–f), so D(III) is palindromic, giving 7 at start of f(III). We also have a(I), a(III), d(III), the last of these giving a 5 at the start of D(III). We also know that f(I) is prime, so f(III) is 2×Square, meaning it must be 722.
Now attack D(III) 57?75. B(II) is not a square factor of D(II) and B(I) is not a square, so B(III) is a square factor of D(III): seek such ending 1, ie, look for divisor 11, 19, 21, 29, 31. Only 57475 works, giving B(III) as 121. Now E+E´+e can’t be I since it must be at least 3000. So, examine it in III. This shows e ends 7 or 8, which means F starts 7 or 8 and so can’t give g(III) as F/2.
Now D=g(A–f) implies g < 20 and g(I) is F/2 so F is even and can only be 24. Taking D(I) as g(A–f) gives a solution with f(I) as 929, D(I) 696, F(I) is 24, g(I) is 12 and d(I) is 91. Now e(I) is the square and is 4624, C(I) is 2472, E(I) is 2732 and c(I) is 7263 = 269×27 (Prime × Cube).
Now e(III) can’t be the cube, from C(III), since the cube could inly be that of 8 and it can’t fit. Thus e(II) is the cube and 1728 co C(II) is 61. Now e(III) could be 541, 547 or 557. Looking at E(III) it can only be 5002 (cube + g) so e(III) is 557, F(III) is 768 and C(III) is 7564. 15408 is 107×144, ie, prime x square.
