what is the sum of all the digits of number (10 ^ 50) 74 A

Question

what is the sum of all the digits of number (10 ^ 50) – 74
A 431
B 436
C 440
D 445
E 449

can you please explain the technique to solve these kind of questions

and hey just a selfish interest

in case u like the question please dont forget to click that kudos button :-)

zeenie · Answer

Hi
The answer is C
This is my approach
If it is 10^4-74
this is 9926
so the number of 9s will be n-2 , as only the last 2 digits will be 26. Rest all digits will be 9.
Hence for n=50, number of 9s = 48 and last 2 digits =26
sum it up
48*9+6+2= 440 !!

Cheers

vd · Answer

Thanks a lot for the good explanation

zeenie · Answer

Glad I could help you. :-D

rpmodi · Answer

I know a very easy method to find solution for these kind of questions just by adding the digits in the answer choice . But finding it difficult to introduce the cocept of "Beejank" here ( vedic math concept) ....I will give it a try ...

for any number Beejank is sum of all the digits in that number , now for any arithmatic operation beejank of left hand side = beejank od right hand side

43*67 = 2881

beejank of 43 (4+3 =7)
beejank of 67 (7+6 = 13 ; 1+3 =4)

perform same operation on beejank 7*4 = 28 (2+8=10 ; 1+0 =1) Beejank of left hand side =1

Beejank of 2881 will also be 1

So Beejank of left hand side = beejank of right hand side

Now in our example we know the final outcome will be

999999..........926 ( another concept of Beejank: for any number containing digit 9 we can just add other digits since beejank of 9 + "any digit/number" = beejank of that number

so beejank of left hand side here is 2+6=8 , in the answer choice we only have one option 440 with beejank of 8 .

I know tomorrow morning when I read my own explanation it might sound confusing ;-)

but I will post it anyways here in hope that someone will grasp this concept or may be google on vedic math to realize the potential of powerful vedic math ....

Never mind ;-)

realized that 431 is also one of the option with Beejank of 8 , but you can take the concept of beejank as it is explained here .

but I will post it anyways here in hope that someone will grasp this concept or may be google on vedic math to realize the potential of powerful vedic math .... Never mind ;-)

realized that 431 is also one of the option with Beejank of 8 , but you can take the concept of beejank as it is explained here .

vd · Answer

Good explanation...i ve read up a lil bit of vedic mathematics and found it to be quiet interesting 
Kudos for u for posting this

rpmodi · Answer

yeah , it infact is a very powerful science , based on 16 sutras ...oh well i dont want to get started with it but its a complete set of rules and u can solve most advanced mathematical concepts with it . I only know 0.01% of it :-)

)

chronic · Answer

I got C - 440.

Basically tried to divide 74 from 10000 and 1000000. The number of 9's obtained is 2 less than the number of 0s in 10^n. So 10^50 mean there will be 48 9s. So answer would be - 48*9 + 2 +6

GMATBLACKBELT · Answer

Ok. Here is how I solved. 100-74= 26. 1000-74=926.. Its just going to be 99999999...26 from here on.

Thus 48*9 + 2+6 = 440.

what is the sum of all the digits of number (10 ^ 50) 74 A

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