using numbers 1-5 to make 2 and 4

[debbie2503]

I have to use the numbers 1 2 3 4 5 and the symbols + and - to end in the answer 2 - any ideas anyone, I've been trying for ages?

Also have to use the same numbers and symbols to answer 4 - help again?

Thanks

 

[mmm4444bot]

There is a big difference between "numbers" and "digits".

I mean, for example, you can't use 1 and 5 to write the number fifteen, yes?

The only numbers available are the first five Natural numbers.

When you say to use the symbols - and + you actually mean that you can use only the operations of addition and subtraction, yes?

I mean, for example, you can't use the symbol - as an abbreviation for the factor -1 and write -5, correct?

Are you allowed to use grouping symbols ?

 

[debbie2503]

you can use the numbers combined eg 1 and 5 could be 15 as long as they are all only used once.

The symbols - and + have to be used at least once

 

[lookagain]

Post Deleted

 

[mmm4444bot]

Okay. So we need to use the digits 1, 2, 3, 4, and 5 once each to form Whole numbers.

We need to use the symbol + at least once.

We need to use the symbol - at least once.

We can not use any other symbols, yes? (No grouping symbols)

We can not use the - symbol as a negation (No operations other than addition and subtraction)

Am I correctly interpreting the exercise, now ?

 

[soroban]

Hello, debbie2503!

I have to use the numbers 1 2 3 4 5 and the symbols + and - to end in the answer 2.

Also have to use the same numbers and symbols to answer 4.

If parentheses are allowed:

. . . . \(\displaystyle 13 - (2 + 4 + 5) \;=\;2\)

. . . . \(\displaystyle 14 - (2 + 3 + 5) \;=\;4\)

 

[lookagain]

Re:

We can not use any other symbols, yes? (No grouping symbols)

We can not use the - symbol as a negation (No operations other than addition and subtraction)

It cannot be done without using grouping symbols when the symbol for negation is not allowed.

\(\displaystyle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Number \ of \ orderings:\)
.....................................---------------------------------

\(\displaystyle 13 - (2 + 4 + 5) = 2\) . . . . . . . . . . . . . 6

\(\displaystyle 14 - (2 + 3 + 5) = 4\) . . . . . . . . . . . . . 6

\(\displaystyle 41 - (35 + 2) = 4\) . . . . . . . . . . . . . . . 2

\(\displaystyle 23 - (15 + 4) = 4\) . . . . . . . . . . . . . . . 2

\(\displaystyle 31 - (24 + 5) = 2\) . . . . . . . . . . . . . . . 2

\(\displaystyle 41 - (32 + 5) = 4\) . . . . . . . . . . . . . . . 2

\(\displaystyle 23 - (14 + 5) = 4\) . . . . . . . . . . . . . . . 2

\(\displaystyle 31 - (25 + 4) = 2\) . . . . . . . . . . . . . . . 2

Also:

\(\displaystyle 13 - (2 + 4) - 5 = 2\) . . . . . . . . . . . . . 4

\(\displaystyle 13 - (2 + 5) - 4 = 2\) . . . . . . . . . . . . . 4

\(\displaystyle 13 - (4 + 5) - 2 = 2\) . . . . . . . . . . . . . 4

\(\displaystyle 14 - (2 + 3) - 5 = 4\) . . . . . . . . . . . . . 4

\(\displaystyle 14 - (2 + 5) - 3 = 4\) . . . . . . . . . . . . . 4

\(\displaystyle 14 - (3 + 5) - 2 = 4\) . . . . . . . . . . . . . 4

\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\


In the last six examples, there are indicated four orderings apiece, as in these for instance:

\(\displaystyle 13 - (2 + 5) - 4 =\)

\(\displaystyle 13 - (5 + 2) - 4 =\)

\(\displaystyle 13 - 4 - (2 + 5) =\)

\(\displaystyle 13 - 4 - (5 + 2)\)

For instance, when looking at certain approaches, you can't have all of the digits separated
(in any order), with no grouping symbols, and be able to get an even result, because the
three odd digits (representing odd numbers there in that particular case) keep the result
from being even.