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The sum of all the digits of the positive integer q is equal

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Re: The sum of all the digits of the positive integer q is equal  [#permalink]

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New post 06 Nov 2017, 02:30
Lets say q= 10^24-49
ie 1000......zero 24 times -49
consider 1000-49 = 951
means when subtracted
9999...22 times then 51 = 9999.....9951 where 9 are 22 times => consider sum => 9x22 +5+1 => units digit is 9x2 +6 => 8+6=4
option a - gives 9x2+6= 24=4
option b - 9x3+6 =33=3 BINGO as we need in format x13 => means we required units place =3
option c = 2 units digit
d = 1 units digit
e = 0 units digit
-----
ANSWER =B
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Re: The sum of all the digits of the positive integer q is equal  [#permalink]

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New post 26 Jun 2018, 08:57
[quote="Bunuel"][quote="farhanabad"]How do you get X = 2?

My approach was after finding out 51, I started plugging in the numbers from the answers and all I was looking for is a number that sums up to a units digit of 3.

[quote="Bunuel"][quote="enigma123"]The sum of all the digits of the positive integer q is equal to the three-digit number x13. If q = 10^n – 49, what is the value of n?
(A) 24
(B) 25
(C) 26
(D) 27
(E) 28

If u plz Bunnel
I have an other approach:
we have 23*9=207
207+5+1=213
is it right?
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Re: The sum of all the digits of the positive integer q is equal  [#permalink]

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New post 01 Dec 2019, 03:31
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Re: The sum of all the digits of the positive integer q is equal   [#permalink] 01 Dec 2019, 03:31

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