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Let S(n) equal the sum of the digits of positive integer n. For exampl

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Let S(n) equal the sum of the digits of positive integer n. For exampl  [#permalink]

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New post 03 Apr 2019, 04:19
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Let S(n) equal the sum of the digits of positive integer n. For example, S(1507) = 13. For a particular positive integer n, S(n) = 1274. Which of the following could be the value of S(n + 1)?

A) 1

(B) 3

(C) 12

(D) 1239

(E) 1265

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Re: Let S(n) equal the sum of the digits of positive integer n. For exampl  [#permalink]

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New post 03 Apr 2019, 11:24
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Noshad wrote:
Let S(n) equal the sum of the digits of positive integer n. For example, S(1507) = 13. For a particular positive integer n, S(n) = 1274. Which of the following could be the value of S(n + 1)?

A) 1

(B) 3

(C) 12

(D) 1239

(E) 1265


When we add digits, the sum of digit repeteadly added till it becomes a single digit will keep increasing by 1..
For example..
2456 .. 2+4+5+6=17....1+7=8
Next number 2457...2+4+5+7=18...1+8=9.. So, the sum increased from 8 to 9.

Let us apply the same on this question..
S(n) = 1274...1+2+7+4=14...1+4=5. Thus, S(n+1) should have a total of 5+1=6 when repeatedly added till it becomes a single digit.

Let us look at the choice which fits in..

A) 1 ....1..NO

(B) 3 .....3...NO

(C) 12 .....1+2=3...NO

(D) 1239 ....1+2+3+9=15...1+5=6...YES

(E) 1265.....1+2+6+5=14...1+4=5...NO

Only D fits in

D
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Let S(n) equal the sum of the digits of positive integer n. For exampl  [#permalink]

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New post 04 Apr 2019, 10:06
chetan2u wrote:
Noshad wrote:
Let S(n) equal the sum of the digits of positive integer n. For example, S(1507) = 13. For a particular positive integer n, S(n) = 1274. Which of the following could be the value of S(n + 1)?

A) 1

(B) 3

(C) 12

(D) 1239

(E) 1265


When we add digits, the sum of digit repeteadly added till it becomes a single digit will keep increasing by 1..
For example..
2456 .. 2+4+5+6=17....1+7=8
Next number 2457...2+4+5+7=18...1+8=9.. So, the sum increased from 8 to 9.

Let us apply the same on this question..
S(n) = 1274...1+2+7+4=14...1+4=5. Thus, S(n+1) should have a total of 5+1=6 when repeatedly added till it becomes a single digit.

Let us look at the choice which fits in..

A) 1 ....1..NO

(B) 3 .....3...NO

(C) 12 .....1+2=3...NO

(D) 1239 ....1+2+3+9=15...1+5=6...YES

(E) 1265.....1+2+6+5=14...1+4=5...NO

Only D fits in

D


Hello chetan2u

nice explanation.
I have one doubt. how do we know that eventually, we have to add the digits of a result which is derived from adding digit of a different number?
S(1239)=15 .... S(15)=1+5=6 -- How do I know that I have to further add digits of the number 15.
Please explain.

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Re: Let S(n) equal the sum of the digits of positive integer n. For exampl  [#permalink]

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New post 11 Apr 2019, 08:10
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chetan2u wrote:
Noshad wrote:
Let S(n) equal the sum of the digits of positive integer n. For example, S(1507) = 13. For a particular positive integer n, S(n) = 1274. Which of the following could be the value of S(n + 1)?

A) 1

(B) 3

(C) 12

(D) 1239

(E) 1265


When we add digits, the sum of digit repeteadly added till it becomes a single digit will keep increasing by 1..
For example..
2456 .. 2+4+5+6=17....1+7=8
Next number 2457...2+4+5+7=18...1+8=9.. So, the sum increased from 8 to 9.

Let us apply the same on this question..
S(n) = 1274...1+2+7+4=14...1+4=5. Thus, S(n+1) should have a total of 5+1=6 when repeatedly added till it becomes a single digit.

Let us look at the choice which fits in..

A) 1 ....1..NO

(B) 3 .....3...NO

(C) 12 .....1+2=3...NO

(D) 1239 ....1+2+3+9=15...1+5=6...YES

(E) 1265.....1+2+6+5=14...1+4=5...NO

Only D fits in

D


It's a very good logic! Brilliant!
But it doesn't strike when needed. Is this the only way to solve this question?
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Re: Let S(n) equal the sum of the digits of positive integer n. For exampl  [#permalink]

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New post 11 Apr 2019, 17:05
S(n+1) for most numbers will just be S(n) + 1. But if there is a 9 at the end, the number "resets". As none of the answer choices is 1274 +1, the number must have a 9 at the end. You can try a few experiments to find a pattern

S(9)=9; S(10)=1; Difference = 8
S(29)=11; S(30)=3; Difference = 8
S(99)=18; S(100)=1; Difference = 17
S(999)=27; S(1000)=1; Difference = 26

So the difference is always 1 less than a multiple of 9. The answer has to be in the form 1274 - (9I -1) = 1275 - 9I where I is an integer. Out of all the answer choices, only D can be written in this form (1275 - 1239 = 36 = 9*4)
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Let S(n) equal the sum of the digits of positive integer n. For exampl  [#permalink]

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New post 28 Jun 2019, 05:29
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Here you can assume that the number of 9's present in S(n) would be the quotient of (1274 / 9) and the reminder would be 5, right?

So, for example, if you have this number 59999999990999999999...9999999, the sum of digits would be 1274, right? And if one adds +1 to this number the sum of digits would be:
1274 - 9K (because of the 9's turning to 0) + 1 (because of the that one 9 that adds 1 to the next digit)

the result is an answer that matches 1275 - 9k, so D is our answer.

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Re: Let S(n) equal the sum of the digits of positive integer n. For exampl  [#permalink]

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New post 28 Jun 2019, 07:21
When you sum any positive integer's digits, then the remainder you get when you divide that sum by 9 will equal the remainder when you divide your original number by 9. So in this case, if our digits sum to 1274, the remainder when we divide n by 9 will be 5, because that's the remainder we get when we divide 1274 by 9 (which you can find by summing the digits of 1274 to get 14). If the remainder is 5 when we divide n by 9, the remainder must be 6 when we divide n+1 by 9. So we just want an answer with that remainder, and 1239 is the only answer choice with a remainder of 6 when we divide it by 9.
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Re: Let S(n) equal the sum of the digits of positive integer n. For exampl   [#permalink] 28 Jun 2019, 07:21
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