A CrossNumber puzzle with a twist

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
2
down vote

favorite












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



Enjoy!






mathematics crosswords







share|improve this question



















  • 1




    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?
    – Cameron Aavik
    1 hour ago







  • 1




    @CameronAavik It means that $A<B<C<D<cdots<K$
    – TheSimpliFire
    1 hour ago














up vote
2
down vote

favorite












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



Enjoy!






mathematics crosswords







share|improve this question



















  • 1




    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?
    – Cameron Aavik
    1 hour ago







  • 1




    @CameronAavik It means that $A<B<C<D<cdots<K$
    – TheSimpliFire
    1 hour ago












up vote
2
down vote

favorite









up vote
2
down vote

favorite











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



Enjoy!






mathematics crosswords







share|improve this question















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



Enjoy!






mathematics crosswords




mathematics crosswords






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 1 hour ago

























asked 1 hour ago









TheSimpliFire

1,885426




1,885426







  • 1




    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?
    – Cameron Aavik
    1 hour ago







  • 1




    @CameronAavik It means that $A<B<C<D<cdots<K$
    – TheSimpliFire
    1 hour ago












  • 1




    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?
    – Cameron Aavik
    1 hour ago







  • 1




    @CameronAavik It means that $A<B<C<D<cdots<K$
    – TheSimpliFire
    1 hour ago







1




1




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?
– Cameron Aavik
1 hour ago





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?
– Cameron Aavik
1 hour ago





1




1




@CameronAavik It means that $A<B<C<D<cdots<K$
– TheSimpliFire
1 hour ago




@CameronAavik It means that $A<B<C<D<cdots<K$
– TheSimpliFire
1 hour ago










2 Answers
2






active

oldest

votes

















up vote
3
down vote



accepted










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







share|improve this answer






















  • (+1) Congrats! Very neat working :)
    – TheSimpliFire
    15 mins ago











  • +1 Looks Like that is correct
    – Khushraj Rathod
    14 mins ago

















up vote
1
down vote













Partial:




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.






share|improve this answer






















  • I'm not saying your answer is wrong/correct, but what about $H=676$?
    – TheSimpliFire
    1 hour ago










  • 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..
    – Shahriar Mahmud Sajid
    1 hour ago










  • No, there aren't any single digit numbers.
    – TheSimpliFire
    1 hour ago










  • in that case what does the bold lines represent ?
    – Shahriar Mahmud Sajid
    1 hour ago










  • 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
    – TheSimpliFire
    1 hour ago











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
3
down vote



accepted










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







share|improve this answer






















  • (+1) Congrats! Very neat working :)
    – TheSimpliFire
    15 mins ago











  • +1 Looks Like that is correct
    – Khushraj Rathod
    14 mins ago














up vote
3
down vote



accepted










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







share|improve this answer






















  • (+1) Congrats! Very neat working :)
    – TheSimpliFire
    15 mins ago











  • +1 Looks Like that is correct
    – Khushraj Rathod
    14 mins ago












up vote
3
down vote



accepted







up vote
3
down vote



accepted






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







share|improve this answer














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








share|improve this answer














share|improve this answer



share|improve this answer








edited 13 mins ago

























answered 16 mins ago









Cameron Aavik

3165




3165











  • (+1) Congrats! Very neat working :)
    – TheSimpliFire
    15 mins ago











  • +1 Looks Like that is correct
    – Khushraj Rathod
    14 mins ago
















  • (+1) Congrats! Very neat working :)
    – TheSimpliFire
    15 mins ago











  • +1 Looks Like that is correct
    – Khushraj Rathod
    14 mins ago















(+1) Congrats! Very neat working :)
– TheSimpliFire
15 mins ago





(+1) Congrats! Very neat working :)
– TheSimpliFire
15 mins ago













+1 Looks Like that is correct
– Khushraj Rathod
14 mins ago




+1 Looks Like that is correct
– Khushraj Rathod
14 mins ago










up vote
1
down vote













Partial:




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.






share|improve this answer






















  • I'm not saying your answer is wrong/correct, but what about $H=676$?
    – TheSimpliFire
    1 hour ago










  • 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..
    – Shahriar Mahmud Sajid
    1 hour ago










  • No, there aren't any single digit numbers.
    – TheSimpliFire
    1 hour ago










  • in that case what does the bold lines represent ?
    – Shahriar Mahmud Sajid
    1 hour ago










  • 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
    – TheSimpliFire
    1 hour ago















up vote
1
down vote













Partial:




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.






share|improve this answer






















  • I'm not saying your answer is wrong/correct, but what about $H=676$?
    – TheSimpliFire
    1 hour ago










  • 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..
    – Shahriar Mahmud Sajid
    1 hour ago










  • No, there aren't any single digit numbers.
    – TheSimpliFire
    1 hour ago










  • in that case what does the bold lines represent ?
    – Shahriar Mahmud Sajid
    1 hour ago










  • 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
    – TheSimpliFire
    1 hour ago













up vote
1
down vote










up vote
1
down vote









Partial:




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.






share|improve this answer














Partial:




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.







share|improve this answer














share|improve this answer



share|improve this answer








edited 43 mins ago

























answered 1 hour ago









Shahriar Mahmud Sajid

3,129528




3,129528











  • I'm not saying your answer is wrong/correct, but what about $H=676$?
    – TheSimpliFire
    1 hour ago










  • 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..
    – Shahriar Mahmud Sajid
    1 hour ago










  • No, there aren't any single digit numbers.
    – TheSimpliFire
    1 hour ago










  • in that case what does the bold lines represent ?
    – Shahriar Mahmud Sajid
    1 hour ago










  • 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
    – TheSimpliFire
    1 hour ago

















  • I'm not saying your answer is wrong/correct, but what about $H=676$?
    – TheSimpliFire
    1 hour ago










  • 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..
    – Shahriar Mahmud Sajid
    1 hour ago










  • No, there aren't any single digit numbers.
    – TheSimpliFire
    1 hour ago










  • in that case what does the bold lines represent ?
    – Shahriar Mahmud Sajid
    1 hour ago










  • 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
    – TheSimpliFire
    1 hour ago
















I'm not saying your answer is wrong/correct, but what about $H=676$?
– TheSimpliFire
1 hour ago




I'm not saying your answer is wrong/correct, but what about $H=676$?
– TheSimpliFire
1 hour ago












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..
– Shahriar Mahmud Sajid
1 hour ago




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..
– Shahriar Mahmud Sajid
1 hour ago












No, there aren't any single digit numbers.
– TheSimpliFire
1 hour ago




No, there aren't any single digit numbers.
– TheSimpliFire
1 hour ago












in that case what does the bold lines represent ?
– Shahriar Mahmud Sajid
1 hour ago




in that case what does the bold lines represent ?
– Shahriar Mahmud Sajid
1 hour ago












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
– TheSimpliFire
1 hour ago





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
– TheSimpliFire
1 hour ago


















 

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