Aaargh! Does not compute! Brain Teaser

[deptfordmarmoset]

Can you only use each number once? Because if that's right and the last missing number is 76, you've only got 1,2,3,4,5,8,9 left to play with. 7 numbers but 8 remaining places???? Tell you what, in the absence of popcorn, I'm off for a quick sandwich and then I'm meeting a friend in a pub.

 

[Archie_tect]

Er, no DM, you place one of the numbers 1 to 9 in each blank square to make the running total 66 at the end... so the number you start with then is affected by the addition, multiplication, division or subtraction of the remaining 8 left leaving the answer at the end as 66.
...or you wait for the solution, here have some more popcorn.

 

[nickyboy]

Well I managed the "Cheryl's Birthday" question that apparently puts me on a par with a Singaporean 14 year old. But the matrix question......I can see that you work backwards and only certain number combinations work but it's beyond me. For now

 

[Archie_tect]

I got it to work with two 3s.... [pointless post No.35567]

 

[mr messy]

I've given the blanks A-I designations. So now i've got
A B C D E F G H I



1 2 3 4 5 6 7 8 9

Just need to draw the lines between now....

 

[raleighnut]

Hic, who cares Its been a long day

 

[martint235]

Done. Can we have a hard one next time please?

 

[deptfordmarmoset]

Er, no DM, you place one of the numbers 1 to 9 in each blank square to make the running total 66 at the end... so the number you start with then is affected by the addition, multiplication, division or subtraction of the remaining 8 left leaving the answer at the end as 66.
...or you wait for the solution, here have some more popcorn.

Friend didn't turn up so had a pint of Gipsy Hill Dissident and then a Northern Star. Anyhow, from what I remember, the final number is 66. The previous calculation is -10. So, to get 66, the number in the last empty box has to be 76. That's 2 of your numbers used up. 7 numbers, 8 remaining boxes....

 

[classic33]

Done. Can we have a hard one next time please?

Howabout this one?

 

[Tin Pot]

Why do they do this??
Independent: Move over Cheryl's birthday, meet the latest Maths puzzle to scramble your brain. http://google.com/newsstand/s/CBIw1pXm9R8
View attachment 89319
.The idea behind it is simple enough. All you need to do is place the numbers 1 to 9 into the grid in such an order that all of the operations work to offer the magic answer of 66 by the time you reach the end. Give it a go. Oh, and if you haven’t worked the funny colon symbol out, that’s meant to be division.

What's the answer??

Hmm.

I'm close, 56.66

 

[Tin Pot]

A +13 *B \C
A should be odd, B&C even
+D +12 x E
D odd, E could be 1 for simplicity
-F -11 +G xH
Needs to be Even at this point, so H & I even
/I -10 =66

Im at 134. Shuffling the numbers could take some time so Im looking fir a few more logical pieces.

 

[nickyboy]

A +13 *B \C
A should be odd, B&C even
+D +12 x E
D odd, E could be 1 for simplicity
-F -11 +G xH
Needs to be Even at this point, so H & I even
/I -10 =66

Im at 134. Shuffling the numbers could take some time so Im looking fir a few more logical pieces.

I took a similar approach. However, the combinations of H and I don't need to be even. If I is 1, H can be any number from 2-9. Equally, for example, if H is 6 I could be 3 or 2. What you can say about H and I is that H/I must be an integer which removes some combinations (but also leaves a lot!)

I must be missing something in the logic. Going to watch a cricket match this afternoon so will give it some though then

Edit: slight brainwave that might help. Whenever you have an x/y combination that must be an integer. So the certain numbers can't be part of the x/y combinations. For example, 5 and 7 can't be used. So 5 and 7 must be a "+" or "-" numbers. Equally, 6,8,9 can't be denominator There are only three +/- numbers so we have two of them. On the right track I think........

 

[threebikesmcginty]

I couldn't be arsed, in fact I could barely be arsed to type I couldn't be arsed.

 

[Tin Pot]

68! Damn!

Calculators are cheating, so i may have dropped a 2 somewhere

5,2,9
4
1,3
51
7,6
39
4,8
68

 

[Tin Pot]

I took a similar approach. However, the combinations of H and I don't need to be even. If I is 1, H can be any number from 2-9. Equally, for example, if H is 6 I could be 3 or 2. What you can say about H and I is that H/I must be an integer which removes some combinations (but also leaves a lot!)

I must be missing something in the logic. Going to watch a cricket match this afternoon so will give it some though then

Edit: slight brainwave that might help. Whenever you have an x/y combination that must be an integer. So the certain numbers can't be part of the x/y combinations. For example, 5 and 7 can't be used. So 5 and 7 must be a "+" or "-" numbers. Equally, 6,8,9 can't be denominator There are only three +/- numbers so we have two of them. On the right track I think........

Im estimating logic, not trying to create hard rules so, if 76 is needed after the last divisor, 76 is likely to be the result of 19x4, 38x2 or 152/2 hence H/I as 1,2,4,8 combinations

Working on the assumption of one of the first two, working back A and B need to result in a low number, because 12 is added and then multiplied (multiplier must be low, probably 3).