Using the numbers 1-6 once each and only once. Create an addition problem using 3 box

[Cillix]

Using the numbers 1-6 once each and only once. Create an addition problem using 3 boxes. Must be 2 numbers per box and cannot add any signs/ symbols / etc. The number has to add up to a number with 2 digits that werent used in any of the numbers used to make it.

Box 1 + box 2 = box 3
(Ex. 12 + 34 = 56)

 

[Deleted member 4993]

Using the numbers 1-6 once each and only once. Create an addition problem using 3 boxes. Must be 2 numbers per box and cannot add any signs/ symbols / etc. The number has to add up to a number with 2 digits that werent used in any of the numbers used to make it.

Box 1 + box 2 = box 3
(Ex. 12 + 34 = 56)

Try the numbers given in the example - but change their order. For example, first box can have 21 (instead of 12) and so on....

Please share your work with us.

 

[Ishuda]

Using the numbers 1-6 once each and only once. Create an addition problem using 3 boxes. Must be 2 numbers per box and cannot add any signs/ symbols / etc. The number has to add up to a number with 2 digits that werent used in any of the numbers used to make it.

Box 1 + box 2 = box 3
(Ex. 12 + 34 = 56)

Sorry but
12 + 34 = 46

 

[Ishuda]

Using the numbers 1-6 once each and only once. Create an addition problem using 3 boxes. Must be 2 numbers per box and cannot add any signs/ symbols / etc. The number has to add up to a number with 2 digits that werent used in any of the numbers used to make it.

Box 1 + box 2 = box 3
(Ex. 12 + 34 = 56)

Label the positions:
ab + cd = ef
I really don't know of any way except to go through the cases. Some aren't so bad, i.e. b+d<10 (why?) will eliminate quite a few cases. Others you just chase down; for example f must be 6 which eliminates a few more.

 

[Deleted member 4993]

Label the positions:
ab + cd = ef
I really don't know of any way except to go through the cases. Some aren't so bad, i.e. b+d<10 (why?) will eliminate quite a few cases. Others you just chase down; for example f must be 6 which eliminates a few more.

Why?

28 + 31 = 59

 

[Ishuda]

Why?

28 + 31 = 59

"...Using the numbers 1-6 once each and only once ..."
ab + cd = ef

Certainly neither a nor c can be 6 [either e would be a three digit number (the other of a or c equal to 5) or e would be greater than 7 (not allowed). If either b or d is 6, then the other must be 5 [else f would be either 0 or greater than 6 neither of which is allowed]. So none of a, b, c, and d can be six. What if e were 6. Well, go through the possibilities, i.e. a (or c) must be 5 and c (or a) must be 1 or ... 'taint possible under the conditions given.

 

[Cillix]

It's a 12th grade quantitative reasoning question

I'm just so stumped on how to go about it but it does want you to use each number 1-6 including 1 and 6 once each once and once only. I don't know what to do. I'm trying really hard and I believe some of you. I think it is impossible. This teacher must be Steven freakin Hawkins.

 

[Deleted member 4993]

"...Using the numbers 1-6 once each and only once ..."
ab + cd = ef

Certainly neither a nor c can be 6 [either e would be a three digit number (the other of a or c equal to 5) or e would be greater than 7 (not allowed). If either b or d is 6, then the other must be 5 [else f would be either 0 or greater than 6 neither of which is allowed]. So none of a, b, c, and d can be six. What if e were 6. Well, go through the possibilities, i.e. a (or c) must be 5 and c (or a) must be 1 or ... 'taint possible under the conditions given.

Alright - going to the corner... send Denis as soon as possible...

 

[Ishuda]

I'm just so stumped on how to go about it but it does want you to use each number 1-6 including 1 and 6 once each once and once only. I don't know what to do. I'm trying really hard and I believe some of you. I think it is impossible. This teacher must be Steven freakin Hawkins.

As I said, you just have to go through a set of possibilities which covers everything - see the 'f must be 6 post above' and the (partial) reasoning behind it. Complete doing that. Since f must be six either
( i) b=1 and d=5 [or b=5 and d=1]
or
(ii) b=2 and d=4 [or b=4 and d=2].
(why?) Given those two cases what must a and c be? So you have four more main cases to go through to cover everything.

So let's see some of your reasoning and maybe we can help if you go astray.

EDIT: Fix a dumb mistake; f must be 6, not e. BTW: I don't know if a solution exists, I haven't worked through everything.

 

[Cillix]

As I said, you just have to go through a set of possibilities which covers everything - see the 'e must be 6 post above' and the (partial) reasoning behind it. Complete doing that. Since e must be six either
( i) b=1 and d=5 [or b=5 and d=1]
or
(ii) b=2 and d=4 [or b=4 and d=2].
(why?) Given those two cases what must a and c be? So you have four more main cases to go through to cover everything.

So let's see some of your reasoning and maybe we can help if you go astray.

EDIT: BTW: I don't know if a solution exists, I haven't worked through everything.

I thought of the same thing since that'd be the only way to end with a six. I also tried having a 5+6 to get 11 and maybe that'd help a little but it didn't. When I get back to all the work I did I'll show alot of what I did. I had no success though.