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k = (10^n) - 38, and the sum of every digit of k is 350.

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gamjatang
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k = (10^n) - 38, and the sum of every digit of k is 350. [#permalink] New post   05 Dec 2005, 05:00
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k = (10^n) - 38, and the sum of every digit of k is 350. What is n?

ex) The sum of every digit of 123 is 6, and the sum of every digit of 54 is 9.

(Answer choices not available)



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Re: k = (10^n) - 38, and the sum of every digit of k is 350. [#permalink] New post   Updated on: 05 Dec 2005, 06:39
gamjatang wrote:
k = (10^n) - 38, and the sum of every digit of k is 350. What is n?

ex) The sum of every digit of 123 is 6, and the sum of every digit of 54 is 9.

(Answer choices not available)


n= 41 :wink:

edit: n= 40

Originally posted by laxieqv on 05 Dec 2005, 05:29.
Last edited by laxieqv on 05 Dec 2005, 06:39, edited 1 time in total.
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Re: k = (10^n) - 38, and the sum of every digit of k is 350. [#permalink] New post   05 Dec 2005, 06:38
gamjatang wrote:
k = (10^n) - 38, and the sum of every digit of k is 350. What is n?

ex) The sum of every digit of 123 is 6, and the sum of every digit of 54 is 9.

(Answer choices not available)


10^n-38 always has form of 9..962 ----> the sum of digits = 9*l+6+2= 350 (l>=0 and l is integer) ---->l= 38 --->there're 38 digits 9
---->10^n-38= 999......962 ( there're 38 of 9) ---> 10^n= 10.....0 (there're 40 of 0 ) -----> n=40
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Re: k = (10^n) - 38, and the sum of every digit of k is 350. [#permalink] New post   05 Dec 2005, 10:31
nice laxieqv :)
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Re: k = (10^n) - 38, and the sum of every digit of k is 350. [#permalink] New post   05 Dec 2005, 10:41
n=2 then 100-38=62
n=3 then 1000-38=962
n=4 then 10000-38=9962
n=5 then 100000-38=99962

For each of them SUM=9*(n-2) + 8

350= 8 + 9 (n-2). -> n-3=38 and n=40
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Re: k = (10^n) - 38, and the sum of every digit of k is 350. [#permalink] New post   05 Dec 2005, 23:10
When n = 2 we get K = 62, sum = 8
When n = 3 we get k = 962 Sum = 17
When n=4 we get k = 9962 sum = 26

The sum forms a sequence 8,17,26,...350
This is an AP we need to find which term is 350.
tn=a+(n-1)d, where a is the first term, d is the diff. between the terms and tn is nth term of the series.

350 = 8 + (n-1)9
350 = 8 + 9n - 9
351 = 9n => n = 39
Thus when n = 39, since the first sum 8 happens when n= 2, so we would get sum of the digits to be 350, when n= 40.
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Re: k = (10^n) - 38, and the sum of every digit of k is 350. [#permalink] New post   06 Dec 2005, 00:09
automan wrote:
n=2 then 100-38=62
n=3 then 1000-38=962
n=4 then 10000-38=9962
n=5 then 100000-38=99962

For each of them SUM=9*(n-2) + 8

350= 8 + 9 (n-2). -> n-3=38 and n=40


Nice analysis, simple and concise.

The OA is 40.



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Re: k = (10^n) - 38, and the sum of every digit of k is 350. [#permalink]
06 Dec 2005, 00:09
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