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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