Question

If m=7777.....777 is a 99 digit number and n=999......999 is a 77 digit number, then the sum of digits in the product of m×n

Answer 1

sum of the digits in m=99×7=693
sum of the digits in n=77×9=693
m×n=693×693=4702495

Answer 2

Qⁿ :If m=7777.....777 is a 99 digit number and n=999......999 is a 77 digit number, then the sum of digits in the product of m×n

solⁿ:  m x n = (777...777) (999...999)

              99 digits     77digits

10² - 1 = 99

10³ - 1 = 999

10⁴ - 1 = 9999

10⁷⁷ - 1 = 999..99

(777...777) (10⁷⁷ - 1)

(777...777)(10⁷⁷) - 7777...777

(777...777 0000 ...000) - ( 777...777)

7777...777   00000..00

7777....77(99)

subtracting we get :

(7777)     ( 6)   (99...999)         (222..222)   (3)

76digits   22digits     1digit      76digits     1digit

sum of the digits

(76 x 7) + (1 x 6 ) + (22 x  9)  + (76 x 2)  + (1 x 3)

532 + 6 + 198 + 152 + 3

=891

SO , IF m =  777...777 IS A 99 DIGIT NUMBER AND  n = 999...999 IS A 77 DIGIT NUMBER , THEN THE SUM OF THE DIGITS IN THE PRODUCT OF    m x n  IS 891

solⁿ by aswin