If m = 777....77777 is a 99 digit number and n = 999....99999 is a 77 digit number, then the sum of the digits in the product m*n is :
a) 890
b) 891
c) 892
d) 893
Please provide your reasoning too if possible.
Answers
Answered by
21
hey mate.
here's the solution
here's the solution
Attachments:
ayushnimak:
Thank you so much!!
Answered by
15
Heya Mate
(•) Let's do it the un-done way !
∆ There's an interesting fact about numbers...
Say, there's a number (abcd) where a,b,c,d are digits ! Kay !
Now, if you remember correctly +_+ that is, if you do at all ^^"
# DIVISIBILITY RULE BY "9"
-> A number is divisible by 9 if and only if the sum of its digits is divisible by 9
____________________________________
(lol)
Now, we have ( mn ) = 9 ( 111111 .... 11 )( 7777777... )
=> ( mn ) is divisible by 9
And so, sum of digits is also divisible by 9 but, the only Option divisible by 9 is -> ( B )
=> Sum of digits is 891
(•) Let's do it the un-done way !
∆ There's an interesting fact about numbers...
Say, there's a number (abcd) where a,b,c,d are digits ! Kay !
Now, if you remember correctly +_+ that is, if you do at all ^^"
# DIVISIBILITY RULE BY "9"
-> A number is divisible by 9 if and only if the sum of its digits is divisible by 9
____________________________________
(lol)
Now, we have ( mn ) = 9 ( 111111 .... 11 )( 7777777... )
=> ( mn ) is divisible by 9
And so, sum of digits is also divisible by 9 but, the only Option divisible by 9 is -> ( B )
=> Sum of digits is 891
Oh wow! That is an interesting way to solve it! Thank you so much!
However the number 891 represents the number of digits in the product, rather than the sum of all the digits. So how are you able to confirm that the sum of the digits would also be divisible by 9?
Read your question :p and then read my answer +_+
Oh, Ok. Sorry, my bad!
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