Sum of digits

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[email protected]: Master | Next Rank: 500 Posts; Posts: 429; Joined: Wed Sep 19, 2012 11:38 pmFollowed by: 4 members; Thanked: 6 times

by [email protected] » Sun Nov 24, 2013 7:46 pm

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

Any idea how to approach this problem

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Source: Beat The GMAT — Problem Solving | Sun Nov 24, 2013 7:46 pm

theCodeToGMAT: Legendary Member; Posts: 1556; Joined: Tue Aug 14, 2012 11:18 pmFollowed by: 34 members; Thanked: 448 times; GMAT Score:650

by theCodeToGMAT » Sun Nov 24, 2013 9:42 pm

q = 10^n - 49

n = 2 => 51 == Sum 6
n = 3 => 951 = sum 15
n = 4 => 9951 = Sum 24

Using AP formula
6 + (n-1)9 =
6 + 9n -9
9n - 3 = x13

So, lets try a value which results in unit digit as "6" when multiplied by 9 ..
i.e. 24

==> 9(24) - 3 ==> 216 - 3 = 213
So, answer n+1 = 25
[spoiler]{B}[/spoiler]

R A H U L

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[email protected]: Master | Next Rank: 500 Posts; Posts: 429; Joined: Wed Sep 19, 2012 11:38 pmFollowed by: 4 members; Thanked: 6 times

by [email protected] » Sun Nov 24, 2013 10:11 pm

Thanks Nice approach dint think of it this way another way I have worked out is below

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?

in 10^n the total number of digits will be n+1 i.e. n zeros and 1 for example if 10^2 then its equal to two zeros and 1

Now the given equation is 10^n-49= will give us n-2 9s and 51 as the last two digits for example if its 10000 then we will get 9951;

so the equation (sum) will be 9(n-2)+5+1=X13, 9n-12=x13,

9n=x25, the sum of the digits should be divisible by 9 therefore x=2 and n=25

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fifafreak: Junior | Next Rank: 30 Posts; Posts: 15; Joined: Wed Jan 27, 2010 3:47 am; Thanked: 2 times

by fifafreak » Mon Nov 25, 2013 6:08 am

[email protected] wrote: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

Any idea how to approach this problem

n=2 : 51
n-3 : 951
n=4 : 9951 .. no. of 9's for each n is given by n-2.

So, sum of the digits = 9*(n-2)+5+1 = 9*(n-2) + 6 [since, last digit of the sum is 3(i.e,x13), 9*(n-2) should have 7 as the last digit => n-2 should have 3 as the last digit, which is possible only for [spoiler]n=25[/spoiler]

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Mathsbuddy: Master | Next Rank: 500 Posts; Posts: 447; Joined: Fri Nov 08, 2013 7:25 amFollowed by: 1 member; Thanked: 25 times

by Mathsbuddy » Mon Nov 25, 2013 8:02 am

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

10^n - 49 = 99999...00 + 51 where the number of 9s = P

So adding digits: 9P + 51 = 100x + 13

So 9p + 38 = 100x

so x = (9p + 38)/100

9p must end in digits 62 (as 62 + 38 =100)
The only multiple of 9 ending thusly is 162 (because 1 + 6 + 2 = 9)

Therefore x = 200/100 = 2

Hence q = 213
and p = 162/9 = 18

Hence (18 digits of 9)*100 + 51 = 10^n - 49

So (18 digits of 9)*100 + 100 = 10^n
99999999999999999900+100=100000000000000000000=10^20
so n = 20

Who can find the error?

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akash.delsaria: Junior | Next Rank: 30 Posts; Posts: 13; Joined: Sun Oct 17, 2010 11:07 am; Thanked: 4 times; GMAT Score:700

by akash.delsaria » Tue Nov 26, 2013 4:20 am

Hi Mathsbuddy

In your response,
"10^n - 49 = 99999...00 + 51 where the number of 9s"
is a mathematical equation and is correct.

But because the mathematical equation is correct, you can not equate the sum of digits in the equation in the form
" So adding digits: 9P + 51 = 100x + 13 "

What you are basically doing is this :

1000 - 49 = 900 + 51 ( Which is correct )

But

1 - 13 = 9 + 6 ( The sum of the digits )

These will not match up.

Also, you have made mistakes in taking the sum of digits ( 51 should be 5+1 = 6, and not be taken as 51) and 10^n - 49 does not equate 100x + 13 (change of signs)

So this is where I believe you went wrong. I hope the explanation given by the others is clear and you have understood the solution.

Regards

YNWA

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Mathsbuddy: Master | Next Rank: 500 Posts; Posts: 447; Joined: Fri Nov 08, 2013 7:25 amFollowed by: 1 member; Thanked: 25 times

by Mathsbuddy » Tue Nov 26, 2013 5:35 am

akash.delsaria wrote:Hi Mathsbuddy

In your response,
"10^n - 49 = 99999...00 + 51 where the number of 9s"
is a mathematical equation and is correct.

But because the mathematical equation is correct, you can not equate the sum of digits in the equation in the form
" So adding digits: 9P + 51 = 100x + 13 "

What you are basically doing is this :

1000 - 49 = 900 + 51 ( Which is correct )

But

1 - 13 = 9 + 6 ( The sum of the digits )

These will not match up.

Also, you have made mistakes in taking the sum of digits ( 51 should be 5+1 = 6, and not be taken as 51) and 10^n - 49 does not equate 100x + 13 (change of signs)

So this is where I believe you went wrong. I hope the explanation given by the others is clear and you have understood the solution.

Regards

Thank you Akash,
You did well to decypher my cryptic reasoning and sort it out!

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Mathsbuddy: Master | Next Rank: 500 Posts; Posts: 447; Joined: Fri Nov 08, 2013 7:25 amFollowed by: 1 member; Thanked: 25 times

by Mathsbuddy » Tue Nov 26, 2013 5:36 am

akash.delsaria wrote:Hi Mathsbuddy

In your response,
"10^n - 49 = 99999...00 + 51 where the number of 9s"
is a mathematical equation and is correct.

But because the mathematical equation is correct, you can not equate the sum of digits in the equation in the form
" So adding digits: 9P + 51 = 100x + 13 "

What you are basically doing is this :

1000 - 49 = 900 + 51 ( Which is correct )

But

1 - 13 = 9 + 6 ( The sum of the digits )

These will not match up.

Also, you have made mistakes in taking the sum of digits ( 51 should be 5+1 = 6, and not be taken as 51) and 10^n - 49 does not equate 100x + 13 (change of signs)

So this is where I believe you went wrong. I hope the explanation given by the others is clear and you have understood the solution.

Regards

Thank you Akash,
You did well to decypher my cryptic reasoning and sort it out!

Like this post

Matt@VeritasPrep: GMAT Instructor; Posts: 2630; Joined: Wed Sep 12, 2012 3:32 pmFollowed by: 119 members; Location: East Bay all the way; Thanked: 625 times; GMAT Score:780

by Matt@VeritasPrep » Sun Dec 01, 2013 9:57 pm

An easy way to do this is to test small values of n and look for a pattern, e.g.

10Â² - 49 = 51 = sum of digits of 6
10Â³ - 49 = 951 = sum of digits of 15
10â�´ - 49 = 9951 = sum of digits of 24

At this point we can see the pattern: the sum of the digits is 9 * (n-2) + 5 + 1.

9 * (n-2) + 5 + 1 = 100x + 13

9 * (n-2) = 100x + 7

So the units digit of 9 * (n - 2) is 7, meaning that (n - 2) must end in 3, and that n must end in 5. At this point, just use the answers.

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