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# If x = 10^50 - 57, what is the sum of all the digits of x?

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Re: If x = 10^50 - 57, what is the sum of all the digits of x? [#permalink]
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10^1 = 10
10^2 = 100
..
10^50 = 50 zeroes

1000-57 = 943 -> (3-2)9 + 4 +3
10^50-57 = (50-2)(9) + 4 +3 = 439
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Re: If x = 10^50 - 57, what is the sum of all the digits of x? [#permalink]
hello Bunuel,

can you explain this one also. thanks man
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Re: If x = 10^50 - 57, what is the sum of all the digits of x? [#permalink]

Mansouri wrote:
If x = 10^50 - 57, what is the sum of all the digits of x?

(A) 390
(B) 418
(C) 420
(D) 439
(E) 449­

hello Bunuel,

can you explain this one also. thanks man

­
$$10^{50}$$ has 51 digits: 1 followed by 50 zeros;
$$10^{50}-57$$ has 50 digits: 48 nines and 43 at the end;

So, the sum of the digits of $$10^{50}-74$$ equals to 48*9 + 4 + 3 = 439.

Hope it helps.­­
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Re: If x = 10^50 - 57, what is the sum of all the digits of x? [#permalink]
10^2-57=43
10^3-57=943

10^n-54=999...(n-2) times 43

sum of Digits=9(n-2)+4+3=9(48)+4+3
9(50-2)+7=450-18+7=450-11=439
Re: If x = 10^50 - 57, what is the sum of all the digits of x? [#permalink]
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