Cryptarithmetic Problem - 1
Ques. For the following Cryptarithmetic find the answers to the below questions?
Since, for Row 3 JE x B give JE itself thus B=1. J E 1 1 —– -> J E (Row 3) J E 0 <-———-
B A D E
J E
B B
----
J E
J E A
--------
B A D E
Assume that E = 4
| 1. | Value of 5B? | |||
| (a) 15 | (b) 10 | (c) 5 | ||
| 2. | Value of B + A? | |||
| (a) 7 | (b) 5 | (c) 1 | (d) 8 | |
| 3. | Value of D ? | |||
| (a) 4 | (b) 3 | (c) 7 | (d) 8 | |
Clearly, from second row where x is written in common multiplication that value is not considered so ideally here we will hav the value A = 0.It is very obvious that the value of B is 1 For e.g.
Since, for Row 3 JE x B give JE itself thus B=1. J E 1 1 —– -> J E (Row 3) J E 0 <-———-
B A D E
J E
1 1
----
J E
J E 0
--------
1 0 D E
Now in the hundred's place, J + value = 10. Also, J is non zero since A = 0. Thus J should be 9. When you add something to the single digit number that results in 10. So J = 9. Now, problem looks like-
9 E
1 1
----
9 E
9 E 0
--------
1 0 D E
Unfortunately we can't predict the values of values of E and D, but we can obviously deduce that , that E and D are consecutive and D = 9 + 1(carry over) + E. Thus D = E + 1 and E = 0 + E As it is given that E = 4, we can say D = 3.
No comments:
Post a Comment