Grid trick

[rust8y]

This 2x5 grid is filled with digits which are all different. the 4 digit number B is a multiple of the 4 digit number A. All but one of the 2 digit down numbers are primes. No 2-digit and 4-digit number starts with 0.

 

In which square must 0 be placed? Explain why this is the only possible square.

 

Complete the grid and explain at each step why you placed each digit in a particular position.

[matt grime]

I think you've definitely shaded in one too many yellow squares.

 

It is reasonalby easy to work out what digits must be in each row by the primality constraint, and then to use the multiple constraint.

 

You might want to put these in the puzzles forum, since that is what it is more than a mathematics topic.

[5614]

I don't get it... what is B and A?

 

I know they're both 4 digits and B=xA where x is some number. But where does A come into it? Is it a row? What 4 numbers make up A and where do they come from?

 

No 2-digit and 4-digit number starts with 0.

And by "2 digit down numbers" I assume you mean a cell on the top row and the corresponding one below it?

 

And the 4 digits referred to in the quote, is that A and B?

[matt grime]

Use a little intution, A and B are going to be (in all likelihood) the two 4-digit numbers, and probably A will be at the top and B at the bottom.

[5614]

Oh, one sec, my bad... I didn't really notice the darker lines on the grid seperating the squares into 4 and 1. Sorry.

[rust8y]

0 is definitely in the bottom row. Where is its position and what are the digits in the bottom row?

[matt grime]

Is that a question because you don't know the answer, in which case I'm not going to reply to it because you should try to work it out (the hint I've given is more than enough to work it out: how many two digit primes end in 2,4,6, or 8? Can you think of any other digit that a 2-digit prime cannot end in?) or because you know the answer and want to see if anyone else can do it?

 

 

And your grid still has too many yellow shaded squares in it, which makes me think you don't know the answer, so why not explain what you've managed to work out and why.

[5614]

Why is there one too many yellow squares shaded?

 

There cannot be a 0 in the top row because no 2 digit number can begin with 0.

 

There cannot be a 0 in bottom row, 2nd square because the 4 digit number cannot begin with 0.

 

There cannot be a 0 in bottom row end square because A is a multiple of B. If B ended in a 0 then so must A, which would violate the condition that all numbers are different. It would also violate the fact that you cannot have a 0 on the top line.

 

Which brings us back to the question; In which square must 0 be placed? I can't see what to do with the 3 remaining possibilities.

 

The digits in the bottom row are 5 digits from 0, 1, 3, 5, 7, 9 because 0 is in the bottom row so that number would not be a prime. As all other 2 digit numbers must now be prime the only endings are 1, 3, 5, 7 or 9.

 

I'm still not sure how to finish this off though.

 

[edit] unless you use trial and error applying all the constraints thus ruling out possibilities until you are left with a solution.

[s pepperchin]

I am pretty sure that the yellow squares are correct. It says that none of the two digit numbers can start with zero so that means that top row can't have 0 anywhere. And since the bottom 4 digit number can't begin with 0 the left yellow square should be yellow, the yellow square on the right is yellow because the only way for it to be zero is if the number on top would be multiplied by 5 or 10, or end in a 5. All of these possibilities aren't possible since to have a 5 at the end makes the top number non-prime, multipling by 5 requires the top number to end with an even number (again non-prime), and multipling by 10 can't work because that would make the bottom number 5 digits long. Thats all I have to say for now.

[YT2095]

is Binary allowed?

[doG]

If B ended in a 0 then so must A, which would violate the condition that all numbers are different.

Not true. There are certain multiples of any digit that end in 0. Every 5th multiple of 2, 4 and 6; every other multiple of 5 and every 10th multiple of the other digits. Do not read more into your limitations than they actually state.

[matt grime]

There cannot be a 0 in bottom row end square because A is a multiple of B.

 

that is not what is written in the first post. B is a multiple of A, although I am assuming that A is the top row 4 digit number and B the bottom row

[5614]

that is not what you wrote in your first post. YOu said that B is a multiple of A.

Oh yeah. I'm really not properly concentrating on this and I should be before I post.

 

Not true. There are certain multiples of any digit that end in 0. Every 5th multiple of 2, 4 and 6; every other multiple of 5 and every 10th multiple of the other digits. Do not read more into your limitations than they actually state.

Ignoring (just for simplicity for this response) the fact that I got the multiplication the wrong way round... you are correct, 2 lots of 5 does end in a zero, my point was that any lots of 0 will end in a 0 and so if the first 4 digit number ended in 0 then so must the other (because the other is a multiple) and then you would need two zeros, which you are not allowed. However as I got my multiplication the wrong way the point is mute. Because the first 4 digit number is at the top so cannot have a 0 in it.

 

=====

 

Now as matt grime right pointed out I got the multiplication wrong, which means that you could have a 0 bottom row on the far right.

 

s pepperchin: As the bottom row can contain: 0, 1, 3, 5, 7 or 9, which is 6 possible digits for 5 spaces one of those numbers can go on the top row. Using 5 at the top (as the last digit of the 4 digit number) would be a possible combination.

[doG]

is Binary allowed?

How would you fill the grid with digits that are all different, the first requirement given?

[matt grime]

the yellow square on the right is yellow because the only way for it to be zero is if the number on top would be multiplied by 5 or 10,

 

 

that is false

 

or end in a 5. All of these possibilities aren't possible since to have a 5 at the end makes the top number non-prime,

 

there is no constraint that says the 4 digit numbers are prime.

[doG]

All of these possibilities aren't possible since to have a 5 at the end makes the top number non-prime, multipling by 5 requires the top number to end with an even number (again non-prime), and multipling by 10 can't work because that would make the bottom number 5 digits long.

The requirements don't say anything about the 4 digit number A being prime, only that 4 of the 5 two digit numbers are prime.

[YT2095]

How would you fill the grid with digits that are all different, the first requirement given?

 

hell if I know, I hate maths, but not thinking in base 10 MAY make things a little easier/flexible was my logic behind exploring this idea.

[s pepperchin]

The requirements don't say anything about the 4 digit number A being prime, only that 4 of the 5 two digit numbers are prime.

 

oops I read it pretty fast and mis understood

[watever]

this is really confusing me... i need to get this done & i have no idea wat so ever of n e of the numbers in the 2x5 grid HELP PLEASE!

 

Oh yeah. I'm really not properly concentrating on this and I should be before I post.

 

Ignoring (just for simplicity for this response) the fact that I got the multiplication the wrong way round... you are correct' date=' 2 lots of 5 does end in a zero, my point was that [u']any[/u] lots of 0 will end in a 0 and so if the first 4 digit number ended in 0 then so must the other (because the other is a multiple) and then you would need two zeros, which you are not allowed. However as I got my multiplication the wrong way the point is mute. Because the first 4 digit number is at the top so cannot have a 0 in it.

 

=====

 

Now as matt grime right pointed out I got the multiplication wrong, which means that you could have a 0 bottom row on the far right.

 

s pepperchin: As the bottom row can contain: 0, 1, 3, 5, 7 or 9, which is 6 possible digits for 5 spaces one of those numbers can go on the top row. Using 5 at the top (as the last digit of the 4 digit number) would be a possible combination.

[amosquitoes]

bottom row 19730

top row 48652

[thebrains***]

honestly, i think you are off the track. the 2nd row must have a 9730&1 because all prime numbers end in those numbers, therefore you need them on the 2nd row. the 0 goes in the end square, and the top row numbers are 2468&5. the top number goes into the bottom number so keep multiplying it...a hint...the number multiplys by 2 to get the answer. you cant use an 8 6 or 5 because that will turn it into a 5 digit number...

over and out

[matt grime]

honestly, i think you are off the track. the 2nd row must have a 9730&1

 

which would be exactly the same answer given by the previous poster....

[Wintermute]

Nice- One HUGE problem: both 1 & 2 are prime. The question quotes: All but one of the two digit numbers are prime.

[Wintermute]

My Solution was:

1084|3

2|9756

OR

4523|1

7|9046

 

But as you can see it fell through because of that.

[Wintermute]

Got It.

4513|7

8|9026

Wintermute Out.