### Brain Teasers

# Tricky Maths Symbols

Each symbol represents one and only one of the numbers from 1-9 inclusive.

All the terms are added except the last one, which is a multiplication # * $ = &$.

&$ is a double digit number.

Solve :

\ + > = &$

^ + % = &$

? + > + @ =&$

\ + ^ + & = &$

# + ? + ^ = &$

# MULTIPLIED by $ = &$

All the terms are added except the last one, which is a multiplication # * $ = &$.

&$ is a double digit number.

Solve :

\ + > = &$

^ + % = &$

? + > + @ =&$

\ + ^ + & = &$

# + ? + ^ = &$

# MULTIPLIED by $ = &$

### Hint

Need to use a lot of logical thinking and deduction: For a start: if \ + > are the biggest numbers possible, 9 and 8, the &$ = 17& is non zero so it must be 1

& = 1

### Answer

To check your answer:1 = &

2 = @

3 = #

4 = ?

5 = $

6 = \

7 = %

8= ^

9 = >

6 + 9 = 15

8 + 7 = 15

4 + 9 + 2 = 15

6 + 8 + 1 = 15

3 + 4 + 8 = 15

3*5 = 15

For a complete solution:

\ + > = &$

^ + % = &$

Starting with the sums with only two terms; because two terms added together can only have a maximum of (9+8)= 17 and the first digit of &$ is non-zero, so the minimum is 12

Therefore & is 1

# MULTIPLIED by $ = &$

something multiplied by something else must be between 17 and 11 (from above)

There are only a few combinations, but only one possibility here

2 * 6 = 12

2 * 7 = 14

2 * 8 = 16

3 * 4 = 12

3 * 5 = 15

The second multiplying term is equal to the second digit of the answer = 5

Therefore # is 3 and $ is 5

Filling in what's known:

\ + > = 15

^ + % = 15

? + > + @ = 15

\ + ^ + 1 = 15

3 + ? + ^ = 15

\ + > = 15

^ + % = 15

These symbols must be 6, 7, 8, 9 because otherwise a result of 15 can't be got.

They must be either 9 + 6 or 7 + 8 in some order.

\ + ^ + 1 = 15

\ + ^ = 14 they can't be 9 (because 5 is already known.)

So they must be 8 and 6 in some order.

That also means > and % must be 7 and 9 in some order.

? and @ must be 2 and 4 in some order (They are the only symbols left for the only two numbers), and ? + @ = 6

? + > + @ = 15

That means > must be 9

\ + > = 15

That means \ = 6

\ + ^ = 14

^ is equal to 8

and % = 7

Filling in what's known again:

6+ 9 = 15

8 + 7 = 15

? + 9 + @ = 15

6 + 8 + 1 = 15

3 + ? + 8 = 15

? must be 4

and therfore @ = 2

1 = &

2 = @

3 = #

4 = ?

5 = $

6 = \

7 = %

8= ^

9 = >

6 + 9 = 15

8 + 7 = 15

4 + 9 + 2 = 15

6 + 8 + 1 = 15

3 + 4 + 8 = 15

3*5 = 15

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## Comments

Exalt the greatness of King McCabe. This is a great puzzle. And your explanation of how to get the answer leaves little to question. Top Job from begining to end. --RAYN

yes I agree with Rayneeday. Well copied Cathal, you left nothing out from the original, making it easy to understand, congratulations on your Copy and pasting skills, not even changing the symbols was a stroke of genius.

Are you stalking me now? (Thats your standard response isn't it?) You have no idea how hard I laughed when I saw your comment this morning. Ray actually liked my puzzle. (Yes *MY* puzzle)

I found it so funny when you say I didn't even change the symbols! You idiot.

My estimation of you just grows daily!

I found it so funny when you say I didn't even change the symbols! You idiot.

My estimation of you just grows daily!

It's a good puzzle, but you made an incorrect statement in your solution. You said that only 3*5=15 satisfies the last equation, but 6*2=12 is also possible at this stage. You have to use the 5th equation to rule it out.

Yeah I think you are right! I totally missed that there is still the possibility of 6*2 = 12 (Obviously because I forgot multiplication is commutative!)

I don't know if you worked this out for yourself but I don't think most people bother, they might go through the answer looking for obvious mistakes.

Its still right but I think that would make it a lot harder!

I just remember Mad Ades comment from above now! That was funny him posting that, for anyone who cares check out the forum topic under my name! He was made look very silly!

I don't know if you worked this out for yourself but I don't think most people bother, they might go through the answer looking for obvious mistakes.

Its still right but I think that would make it a lot harder!

I just remember Mad Ades comment from above now! That was funny him posting that, for anyone who cares check out the forum topic under my name! He was made look very silly!

6x2=12 quickly leads to a dead end, as stated above. I flipped a coin, went 3x5, made a mistake, tried 6x2, became certain of the error, then retraced 3x5 and solved the problem.

This is an excellent puzzle - I was close to taking a rest after figuring out I made a mistake, but plowed on for the satisfaction. Sweet! More more!

This is an excellent puzzle - I was close to taking a rest after figuring out I made a mistake, but plowed on for the satisfaction. Sweet! More more!

excellent excellent teaser

Excelent problem! Great job Cath.

Cute!

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