10. Each alphabet stands for one digit in the following
multiplication.
T H I S
x I S
---------
X F X X
X X U X
------------
X X N X X
------------
What is the maximum value T can take?
Answers were Sorted based on User's Feedback
Answer / harshit gupta
max value of t =4;
if t=5 then (5 H I S)* S = XFXX = 4 digit no.
in this case max value of S = 1 cz if we take s>1 (eg. S=2) then XFSS will be 5 digit no..but it is 4 digit no...if we take S=1 then 5 H I S * 1 = 5 H I S not equal to XFXX;
so 5 is not possible.....
if T=4 then S=1 is not possible..but we can take S=2;
(2 H I S) * 2= 4 digit no...
so, max value of T =4.
Is This Answer Correct ? | 1 Yes | 0 No |
Answer / sk
Simple Max value it can take is 9
because they havent given any constraints for I S. So assume I = 1 and S = 1 and no limit for H so H = 1
Now (T 1 1 1) * (1 1) - This can be the case where T will be Max
So Answer is '9' and this will also match the total result of 5 digit (means after adding the sum of each multiply of 9 1 1 1 with 1).
Finally,
9111 (THIS)
*11 (IS)
----
9111 (XFXX)
+9111 (XXUX)
-----
18222 (XXNXX)
PS: They dint mentioned that X should be same.
Is This Answer Correct ? | 2 Yes | 1 No |
How to write AM/FM demodulation in C language ?
Can a generic class extend another generic class??
hai im going to face ibps SO IT officers grade 1 interview. if anyone had an idea plz tel me expected and experienced questions and mail to anusha_k_anu@yahoo.com
what is b tree
Write a c pgm to print the letter as per given condition i.e.. if u give 4 out put should b 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4
what are the differences between structures and arrays in c language?
Whatis Bandwidth?
WHY WE SELECT U AS A PERTOLEUM ENGINEER THOUGH UR MARKS ARE LESS THEN OTHER APPLICANTS
What is a decade?
what is the command used to find length of the document in unix
i want what are the subjects that are in group1 exams. plz send me the subjects.
What is the time complexity T(n) of the nested loops below? For simplicity, you may assume that n is a power of 2. That is, n = 2k for some positive integer k. : i = n; while (i >= 1){ j = i; while (j <= n) { <body of the inner while loop > // Needs (1). j = 2 * j; } i = i/2; } :