Fill in the Empty Boxes
Is it possible to fill each box in with an arithmetic operation so that this becomes a true equation?
Did you too find it true?
Did you too find it true?
Posts
Correct Operators in Empty Boxes!
What wasn't looking possible?
Yes, it's possible. All you need to do is recall BODMAS (Brackets, Of, Division, Multiplication, Addition, Subtraction) rule in mathematics that we learned in school.
Correcting The Incorrect
What was the wrong?
All we need to do is move 'this' pointed match stick.
Now it looks like -
This is now correct equation. Isn't it?
Wrong Looking Correct Mathematical Equation!
The following question it puts forth you:
25 - 55 + (85 + 65) = ?
Then, you are told that even though you might think its wrong, the correct answer is actually 5!
Whats your reaction to it? How can this be true?
That's how it's perfectly correct!
25 - 55 + (85 + 65) = ?
Then, you are told that even though you might think its wrong, the correct answer is actually 5!
Whats your reaction to it? How can this be true?
That's how it's perfectly correct!
That's How Equation is Correct!
Why it was looking wrong?
If you read the data carefully then you will notice '!' attached to number 5 which is being claimed answer. Actually claimed answer is 5! not 5 Read it again...
"Then, you are told that even though you might think its wrong, the correct answer is actually 5!."
Now use of '!' is not limited to the sentences only. In mathematics it's a 'factorial'.
So 5! = 5 x 4 x 3 x 2 x 1 = 120 and 25 - 55 + (85 + 65) = 120 and hence,
25 - 55 + (85 + 65) = 5!
Now doesn't it look the correct equation?
The CryptArithmetic Problem
Can you solve the below alphametic riddle by replacing letters of words by a numbers so that the below equation holds true?
BASE +
BALL
---------
GAMES
----------
Find numbers replaced letters here!
Source
BASE +
BALL
---------
GAMES
----------
Find numbers replaced letters here!
Source
The CryptArithmetic Problem's Solution
What was the problem?
Let's first recall the given equation.
BASE +
BALL
---------
GAMES
----------
We are assuming repeating the numbers are not allowed.
Let's first take last 2 digits operation into consideration i.e. SE + LL = ES or 1ES (carry in 2 digit operation can't exceed 1). For a moment, let's assume no carry generated.
10S + E + 10L + L = 10E + S .....(1)
9 (E - S) = 11L
To satisfy this equation L must be 9 and (E - S) must be equal to 11. But difference between 2 digits can't exceed 9. Hence, SE + LL must have generated carry.So rewriting (1),
10S + E + 10L + L = 100 + 10E + S
9 (E - S) + 100 = 11L
Now if [9 (E - S)] exceeds 99 then L must be greater than 9. But L must be digit from 0 to 9. Hence, [9 (E - S)] must be negative bringing down LHS below 100. Only value of E - S to satisfy the given condition is -5 with L = 5. Or we can say, S - E = 5.
Now, possible pairs for SE are (9,4), (8,3), (7,2), (6,1), (5,0). Out of these only (8,3) is pair that makes equation SE + LL = SE + 55 = 1ES i.e. 83 + 55 = 138. Hence, S = 8 and E = 3.
Replacing letters with numbers that we have got so far.
1---------
BA83 +
BA55
---------
GAM38
----------
Now, M = 2A + 1. Hence, M must be odd number that could be any one among 1,7,9 (since 3 and 5 already used for E and L respectively).
If M = 1, then A = 0 and B must be 5. But L = 5 hence M can't be 1
If M = 7, then A = 3 or A = 8. If A = 3 then B = 1.5 and that's not valid digit. And if A = 8 then it generates carry 1 and 2B + 1 = 8 again leaves B = 3.5 - not a perfect digit.
If M = 9, then A = 4 (A = 9 not possible as M = 9) and B must be 7 with carry G = 1.Hence for first 2 digits we have 74 + 74 + 1 = 149.
Finally, rewriting the entire equation with numbers replacing digits as -
1
---------
7483 +
7455
---------
14938
----------
So numbers for letters are S = 8, E = 3, L = 5, A = 4, B = 7, M = 9 and G = 1.
