## Blog-94

Dear Students,
Today we will see some riddles. Actually the riddles are very simple. Only we need to think on the riddle with the basic concept of the topic.

We all know the decimal system very well. There are 10 digits say 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. In this riddle, we changed all the values of digits 0 to 9. On the basis of these changed values of all these digits, following problems are solved. Study all these problems very carefully.

Let us discuss some steps of the solution of such problems.

Solved Example:

Key-1

Sum of all the digits from 0 to 9 is 58.

Key-2

853
+ 388
-----------
818

Key-3

5381
+ 1573
-------------
26501

Key-4

9
x 6
-----------
75

### Step-1

Using Key 1:

As we know that the sum of all the digits from 0 to 9 is 45 in actual case. Here in the problem, for changed values of the digits, the sum is 58 so original digit for 5 is 4 and that of 8 is 5. This means, for new value 5, the original value of digit is 4 and for new value 8, the original value of digit is 5.

### Step-2

Prepare a chart of New values and the original values of all the digits as shown bellow:

 Original Digits 0 1 2 3 4 5 6 7 8 9 New Values of Digits 5 8
Like this, find out all the values of remaining digits from other keys given in the question.

Using Key 2:

Problem with new values          Problem with original values

853                                            54a             540
+ 388                                         + a55          + 055
---------                                       ----------         ----------
818                                           5b5             595

Here, a + 5 = 5 so, a = 0, so, 3 is 0 and here we have to add 540 and 055 which comes to 595 so b = 9 i.e. 1 is 9.

Now write the original values of 3 as 0 and 1 as 9 in the following table.

 Original Digits 0 1 2 3 4 5 6 7 8 9 New Values of Digits 3 5 8 1

Using Key 3 : From above table, we have,

Problem with new values          Problem with original values

5381                                       4059                4059
+ 1573                                    + 94a0             + 9420
------------                                  ----------            ----------
26501                                     dc4b9              13479

Here , it clear that d = 1 and c = 3 as 4 + 9 = 13 with no carry from previous digits. Secondly a can't be 6, 7 or 8 as no carry is forwarded. So a must be 2, i. e. 7 is 2. As 5 + 2 = 7, our b is 7, i. e. 0 is 7.

Now write the original values of 7 as 2, 0 as 7, 2 as 1 and 6 as 3 in the following table.

 Original Digits 0 1 2 3 4 5 6 7 8 9 New Values of Digits 3 2 7 6 5 8 0 1

Using Key 4 : From above table, we have,

Problem with new values          Problem with original values

9                                       a                     8
x 6                                    x 3                   x 3
-------                             ----------            ----------
75                                     24                   24

Here , it clear that a = 8 as 8 x 3 = 24. So a must be 8, i. e. 9 is 8 and remaining 4 must be 6.

Now write the original values of 9 as 8, and 4 as 6 in the following table.

 Original Digits 0 1 2 3 4 5 6 7 8 9 New Values of Digits 3 2 7 6 5 8 4 0 9 1

Now we can answer all the questions if new value is allotted to any digit.

Using above solved example, we can answer the following:

Key-1

Sum of all the digits from 0 to 9 is 80.

Key-2

801
+ 488
-----------
7248

Key-3

086
+ 167
-----------
954