Here's an interesting puzzle which I discovered the solution to yesterday.

The eleven grid entries in increasing order are $A, B, C, D, E, F, G, H, I, J, K$, and you are given the following hints. Each of $A-K$ is distinct and none of them start with a zero. They are all integers, and you need to work out where $A-K$ go in the CrossNumber diagram below.

  • $B, E, F, G, H, J$ are squares

  • $A, H, K$ are palindromic numbers

  • $C, D$ are primes

  • $B$ is triangular

  • $I$ is Fibonacci

  • $J$ is the reverse of $F$

P.S. The thicker lines are meant to act as breaks.

enter image description here


  • 1
    $\begingroup$ You say "The eleven grid entries in increasing order". What do you mean by this? Increasing order in value? Increasing order in position on the CrossNumber? $\endgroup$ Oct 6 '18 at 8:09
  • 1
    $\begingroup$ @CameronAavik It means that $A<B<C<D<\cdots<K$ $\endgroup$ Oct 6 '18 at 8:09

Here are the values for each number:

$A=33, B=36, C=37, D=47, E=100, F=169, G=441, H=484, I=610, J=961, K=3663$

Here is the resulting CrossNumber when filled out:

Here is my working out:

$B=36$ as it is the only 2 digit number that is a square triangular number
$F \in \{144, 169\}$ as they are the only 3 digit numbers that have a reverse that is also a square 3 digit number.
$J \in \{441, 961\}$ as a reverse of $F$
$H \in \{484, 676\}$ as it must be larger than $F$ and be a palindromic square
$I=610$ as it has to be larger than $H$ and less than $J$, ($484 < I < 961$), and $610$ is the only fibonacci number in that range
$H=484$ as it has to be less than $I$
$J=961$ as it has to be more than $H$
$F=169$ as the reverse of $J$
After this point I had reduced the set of possible values for each number enough to start placing numbers in places to see what happened and got the solution

  • 1
    $\begingroup$ (+1) Congrats! Very neat working :) $\endgroup$ Oct 6 '18 at 9:39
  • 2
    $\begingroup$ +1 Looks Like that is correct $\endgroup$ Oct 6 '18 at 9:39


OP clarified there are 4 two digit numbers, 6 three digit numbers and 1 four digit number. Since they are in order it means-
Two digit- A,B,C,D
Three digit- E,F,G,H,I,J
Four digit- K
Now, since H is a three digit number that is both square and palindrome, three possible candidates are 121,484,676. But it can't be 121 as there need to be three more three digit square numbers (E,F,G) before H.
Hence, $H = 484$ or $676$
Also, B is triangular square number of two digit, only one that fits is 36
So, $B=36$

This is what i have so far.

  • $\begingroup$ I'm not saying your answer is wrong/correct, but what about $H=676$? $\endgroup$ Oct 6 '18 at 8:45
  • $\begingroup$ oh, completely missed that, also my initial assumption was wrong too, looking at image again it seems there are two single digit numbers, editing accordingly.. $\endgroup$ Oct 6 '18 at 8:47
  • $\begingroup$ No, there aren't any single digit numbers. $\endgroup$ Oct 6 '18 at 8:48
  • $\begingroup$ in that case what does the bold lines represent ? $\endgroup$ Oct 6 '18 at 8:49
  • $\begingroup$ So ACROSS: 1 has three digits, 4 has two, 6 has four, 7 has two, 8 has three, and DOWN: 1 has three digits, 2 has three, 3 has three, 4 has two, 5 has three, 6 has two $\endgroup$ Oct 6 '18 at 8:50

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