The Alphamatic Problem

In the addition below, all digits have been replaced by letters. Equal letters represent equal digits and different letters represent different digits.

   ABCABA 

+

   BBDCAA 

+

   ABEABB 

+ 

   ABDBAA

-------------------

 AAFGBDH

What does the complete addition look like in digits? 

Note : Alphamatic in the title is word derived from Alphabets & Mathematics. In such problems numbers are replaced by alphabets. The challenge is to find the number for each alphabet satisfying given mathematics equation.

Step by step answer is here! 

Source 

Answer Of Alphamatic Problem

Here is question!

 First of all let's write down the equation once again.


   ABCABA 

+

   BBDCAA 

+

   ABEABB 

+ 

   ABDBAA

-------------------

 AAFGBDH

We will refer to places in number from left as a first, second, third...sixth instead of tenth, hundredth, thousandth etc. 

First we need to find if the 5 digits of first number itself i.e. ABCAB are carries forwarded from previous place.  

From the addition of variable from first place, we get,

3A + B  = 10A + A ........(1)

B  =  8A              ........(2)

Only numbers satisfying above are A = 1 & B = 8 , but at previous place we have addition of 4 B's. If B = 8, then addition at second place would be 32 with F = 2 & carry 3 which is not equal to A = 1. So A can't be a carry. So we need to modify (2) above as

B + x = 8A .........(2)....Where 'x' is carry forwarded from second place.

If B = 1 or 2 then x = 0 as  at second place we would have F = 4 or 8. In that case, A would be fractional. Some other possible combinations for B, A & x are,

B = 9, x = 3,   8A = 12,

B = 8, x = 3,   8A = 10,

B = 7, x = 2,   8A =  9,

B = 6, x = 2,   8A = 8,

This is the only combination that can make A a whole number. So A = 1, B = 6.

---------------------------------------------------------------------------------------------

From sixth place, we have,

H = 3A + B = 9

---------------------------------------------------------------------------------------------

From fifth place, we have,

D = 2A + 2B  = 14

But it has to be single digit i.e. D = 4 with 1 carry forwarded to next.

---------------------------------------------------------------------------------------------

From fourth place,

B = 2A + B + C + 1  .....1 is carry from last place.

6 = 9 + C

Now C can't be negative hence C + 9 has to be 2 digit number with 6 at last digit.Since addition of 2 single digit numbers never exceeds 18, C + 9 has to be 16.

16 = 9 + C 

gives, C = 7 & carry 1 forwarded to third place.

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What Is () + () + () = 30?

What is () + () + () = 30?

Fill in blanks with 1,3,5,7,9,11,13,15.

You can also repeat the numbers. 

 Here is probable answer! 

Probable Numbers in () + () + () = 30

What's the question?

Addition of 3 odd numbers can't be even number. Here you have to do the same bringing 30 in result. So you need to do some trick. There are number of possible answers given to this question. Let's see what most common answers given.

1. Using factorial !

3! + 9 + 15 = 30.

2. Using decimal points.

11.3 + 15.7 + 3 = 30.

3. Changing base of numbers.

Assuming it in base 9 numbers,

9 + 9 + 9 = 30.

If we assume ! is allowed as in (1) which can be represented in other way,

(3x2x1) + 9 + 15 = 30

Now some body other replace that bracket with simple addition within it like,

(5+7) + 11 + 7 = 30

or

(15-13) + 15 + 13 = 30

This has to be valid as well & number of such possible combinations are valid!
 
Using decimal point is another form of this trick!

(11 + 3/10) + (15 + 7/10) + 3 = 30
 
In every above trick it's assumed that addition, multiplication, division, subtraction allowed within the bracket!
 
For the change of base, I think they would have written it in subscript of 30 otherwise infinite number of answers again. Like, if we assume it in base 5 then,

13 + 11 + 1 = 30.
 
I think, only answer with less questions is leaving first space blank & write,

+ 15 + 15 = 30

where first '+' is unary sign used to represent positive number while second one is operator for addition. Again, this comes with assumption that using only 2 numbers allowed as it's specifically not mentioned in question that you have to use all 3 numbers.

Probably this is the answer expected otherwise we need to call the creator of this for what he/she is expecting!

The Shopping Challenge

At a stationary shop, shopkeeper sells notebook at Rs.15, Pen at Rs.1 & Eraser at Rs.0.25. You are asked to purchase 100 objects & you are given Rs.100 note. 

So how many notebooks,pens & erasers would you buy?

Shopping Challenge

Here is what should you do! 

Taking The Shopping Challenge

What was the problem?

Let N = number of notebooks, P = number of pens & E = number of erasers.

From the given maths we get 2 equations as,

N + P + E = 100................(1)

15N +1P +E/4=100…....(2)
 
From (1),

P = 100 - N - E

From (2),

60N + 4P + E= 400

60N + 4(100-N-E) + E = 400

60N + 400-4N-4E + E = 400

56N - 3E = 0

56N = 3E

N/E = 3/56

From the ratio, we get, N=3, E=56.

Putting these in equation (1),

We get,

P=41.

So we should buy 56 erasers,41 pens & 3 notebooks to satisfy the given condition.