Blog94
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:
Key1
Sum of all the digits from 0 to 9 is 58.
Key2
853
+ 388

818
Key3
5381
+ 1573

26501
Key4
9
x 6

75
Step1
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.
Step2
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.
+ 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.
+ 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.
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:
1) Using following keys, answer the questions asked bellow these keys:
Key1
Sum of all the digits from 0 to 9 is 80.
Key2
801
+ 488

7248
Key3
086
+ 167

954
Now answer the following ?
a) What is the new value of 3?
b) Write the product 245 x 359 with answer in new values.
c) Find the original value of 375.
d) Rewrite the "(809+548)/70" in original values.
a) What is the new value of 3?
b) Write the product 245 x 359 with answer in new values.
c) Find the original value of 375.
d) Rewrite the "(809+548)/70" in original values.