# Fill in numbers from 1 to 9

Rakesh Joshi

Ranch Hand

Posts: 218

posted 10 years ago

Fill in numbers 1, 2, 3, 4, 5, 6, 7, 8 and 9 into ..... to make the equations work.

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- ....(+)....(=).... -

- --- ---------(/) -

- ....(-)....(=).... -

- -------------(=) -

- ....(=)....(*).... -

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[ March 28, 2006: Message edited by: Rakesh Joshi ]

[ March 28, 2006: Message edited by: Rakesh Joshi ]

----------------------

- ....(+)....(=).... -

- --- ---------(/) -

- ....(-)....(=).... -

- -------------(=) -

- ....(=)....(*).... -

----------------------

[ March 28, 2006: Message edited by: Rakesh Joshi ]

[ March 28, 2006: Message edited by: Rakesh Joshi ]

Life is a Game play it.

posted 10 years ago

I translated this to

Solutions are [71][17]8954632 (a-i).

Most Multiplications get greater than 9 (3*4) so I started there.

1*x is smaller than 10 for every x in (1-9), but would lead to

1*x = x and that's impossible.

h*i=g and i*f=c don't have g or c on the left side.

Therefore h,i,f must be at least 2.

Since 2*5 is 10, every multiplication is too big if a 5(or bigger) and no 1 is involved.

2*3 and 2*4 are the only candidates, which is consistent to our problem: i occures two times and is therefore 2.

h and f are 3 or 4.

Therefore c is 6 or 8 and g is 6 or 8.

If f is 3 or 4, we only have 1,7 and 9 left for d.

But e+f=d, therefore d must be greater than f, which might be 7 or 9.

If d would be 7, e would need to be 3 or 4, which is impossible, since f or h are 3 and 4.

d = 9.

e+ (3,4) = 9 => e:={6, 5}, but since g is 6 or c is 6, e=5 and therefore f=4.

For a and b we have 1 and 7 left, which leads concluent to c=8.

But whether a or b are 1 or 7 is not decideable.

To verify my assumption I elegantly skipped the possibility to create a permutation, and let Random poof my assumption:

Solutions are [71][17]8954632 (a-i).

Most Multiplications get greater than 9 (3*4) so I started there.

1*x is smaller than 10 for every x in (1-9), but would lead to

1*x = x and that's impossible.

h*i=g and i*f=c don't have g or c on the left side.

Therefore h,i,f must be at least 2.

Since 2*5 is 10, every multiplication is too big if a 5(or bigger) and no 1 is involved.

2*3 and 2*4 are the only candidates, which is consistent to our problem: i occures two times and is therefore 2.

h and f are 3 or 4.

Therefore c is 6 or 8 and g is 6 or 8.

If f is 3 or 4, we only have 1,7 and 9 left for d.

But e+f=d, therefore d must be greater than f, which might be 7 or 9.

If d would be 7, e would need to be 3 or 4, which is impossible, since f or h are 3 and 4.

d = 9.

e+ (3,4) = 9 => e:={6, 5}, but since g is 6 or c is 6, e=5 and therefore f=4.

For a and b we have 1 and 7 left, which leads concluent to c=8.

But whether a or b are 1 or 7 is not decideable.

To verify my assumption I elegantly skipped the possibility to create a permutation, and let Random poof my assumption:

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