The sum of all the digits of the positive integer q is equal : Problem Solving (PS)

enigma123

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21 Jan 2012, 17:03

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Sorry Bunuel - Struggling to understand how you got to 10^n-49 will have n digits: n-2 9's and 51 in the end. For example: 10^4=10,000-51=9,949 --> 4 digits: 4-2=two 9's and 49 in the end;

Bunuel

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21 Jan 2012, 17:14

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enigma123

Sorry Bunuel - Struggling to understand how you got to 10^n-49 will have n digits: n-2 9's and 51 in the end. For example: 10^4=10,000-51=9,949 --> 4 digits: 4-2=two 9's and 49 in the end;

\(10^n-49\) will have \(n\) digits: \(n-2\) 9's and 51 in the and. For example: 10^4-49=10,000-49=9,951 --> 4 digits: 4-2=two 9's and 51 in the end;

Or maybe this will help:
10^n has n+1 digits: 1 and n zeros;
10^n-49 has n digits, so one less digit than 10^n. In the end it'll have 51 (the same way 1,000-49=951) and the rest of the digits, so n-2 digits, will be 9's.

Hope it's clear.

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17 Aug 2012, 23:24

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Again the best solution by Bunuel.
I think the question should be written as clear as possible to avoid any confusion.

The stem of the question should be {The sum of all the digits of the positive integer q is equal to the three-digit number which is x13.....} rather than {The sum of all the digits of the positive integer q is equal to the three-digit number x13.}
(A) 24
(B) 25
(C) 26
(D) 27
(E) 28

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18 Jul 2013, 11:59

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I also solved it by finding a pattern:

For n = 3: q = 10^3 - 49 = 951
For n = 4: q = 10^4 - 49 = 9951
For n = 5: q = 10^5 - 49 = 99951
etc...

We can see that the number of 9 digits is 2 less than n. We can now see that our number q equals 99999.....951.
We can now test with the given values for n:

(A) n = 24: 9.(n-2) + (5+1) = 9.(n-2) + 6 = 9(22) + 6 = 204
(B) n = 25: (9)(23) + 6 = 213, our answer is (B)

I think this can be solved in under 2 minutes with this method if one can find the pattern quickly.

farhanabad

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11 Jun 2014, 20:15

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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.

Bunuel

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

Any idea how to approach this problem?

\(10^n\) has \(n+1\) digits: 1 and \(n\) zeros. For example: 10^2=100 --> 3 digits: 1 and 2 zeros;

\(10^n-49\) will have \(n\) digits: \(n-2\) 9's and 51 in the and. For example: 10^4-49=10,000-49=9,951 --> 4 digits: 4-2=two 9's and 51 in the end;

We are told that the sum of all the digits of \(10^n-49\) is equal to the three-digit number \(x13\) --> \(9(n-2)+5+1=x13\) --> \(9n-12=x13\) --> \(9n=x25\) --> \(x25\) is divisible by 9 --> the sum of its digits must be divisible by 9 --> \(x=2\) --> \(9n=225\) --> \(n=25\).

Answer: B.

Hope it's clear.

Bunuel

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12 Jun 2014, 04:56

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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.

Bunuel

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

Any idea how to approach this problem?

\(10^n\) has \(n+1\) digits: 1 and \(n\) zeros. For example: 10^2=100 --> 3 digits: 1 and 2 zeros;

\(10^n-49\) will have \(n\) digits: \(n-2\) 9's and 51 in the and. For example: 10^4-49=10,000-49=9,951 --> 4 digits: 4-2=two 9's and 51 in the end;

We are told that the sum of all the digits of \(10^n-49\) is equal to the three-digit number \(x13\) --> \(9(n-2)+5+1=x13\) --> \(9n-12=x13\) --> \(9n=x25\) --> \(x25\) is divisible by 9 --> the sum of its digits must be divisible by 9 --> \(x=2\) --> \(9n=225\) --> \(n=25\).

Answer: B.

Hope it's clear.

We got that the sum of three-digit number x25 is divisible by 9. A number is divisible by 9 if the sum of its digit is divisible by 9. So, for x25 to be divisible by 9 x must be 2: 2+2+5=9, which is divisible by 9.

sukriti201

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30 Nov 2014, 09:25

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Bunuel

We are told that the sum of all the digits of \(10^n-49\) is equal to the three-digit number \(x13\) --> \(9(n-2)+5+1=x13\) --> \(9n-12=x13\) --> \(9n=x25\) --> \(x25\) is divisible by 9 --> the sum of its digits must be divisible by 9 --> \(x=2\) --> \(9n=225\) --> \(n=25\).

Answer: B.

Hope it's clear.

how come 9n-12 = x13 ---> 9n = x25 ??

Bunuel

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01 Dec 2014, 03:14

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sukriti201

Bunuel

We are told that the sum of all the digits of \(10^n-49\) is equal to the three-digit number \(x13\) --> \(9(n-2)+5+1=x13\) --> \(9n-12=x13\) --> \(9n=x25\) --> \(x25\) is divisible by 9 --> the sum of its digits must be divisible by 9 --> \(x=2\) --> \(9n=225\) --> \(n=25\).

Answer: B.

Hope it's clear.

how come 9n-12 = x13 ---> 9n = x25 ??

Three-digit number x13 plus 12 is three-digit number x25. For example, 113 + 12 = 125.

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26 Dec 2016, 03:45

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Bunuel

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

Any idea how to approach this problem?

\(10^n\) has \(n+1\) digits: 1 and \(n\) zeros. For example: 10^2=100 --> 3 digits: 1 and 2 zeros;

\(10^n-49\) will have \(n\) digits: \(n-2\) 9's and 51 in the and. For example: 10^4-49=10,000-49=9,951 --> 4 digits: 4-2=two 9's and 51 in the end;

We are told that the sum of all the digits of \(10^n-49\) is equal to the three-digit number \(x13\) --> \(9(n-2)+5+1=x13\) --> \(9n-12=x13\) --> \(9n=x25\) --> \(x25\) is divisible by 9 --> the sum of its digits must be divisible by 9 --> \(x=2\) --> \(9n=225\) --> \(n=25\).

Answer: B.

Hope it's clear.

Got confused with this step "x139n−12=x13 --> 9n=x259n=x25 -->" but got to know it's logic in the post mentioned at the bottom.

Bunuel

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26 Dec 2016, 04:02

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anairamitch1804

Bunuel

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

Any idea how to approach this problem?

\(10^n\) has \(n+1\) digits: 1 and \(n\) zeros. For example: 10^2=100 --> 3 digits: 1 and 2 zeros;

\(10^n-49\) will have \(n\) digits: \(n-2\) 9's and 51 in the and. For example: 10^4-49=10,000-49=9,951 --> 4 digits: 4-2=two 9's and 51 in the end;

We are told that the sum of all the digits of \(10^n-49\) is equal to the three-digit number \(x13\) --> \(9(n-2)+5+1=x13\) --> \(9n-12=x13\) --> \(9n=x25\) --> \(x25\) is divisible by 9 --> the sum of its digits must be divisible by 9 --> \(x=2\) --> \(9n=225\) --> \(n=25\).

Answer: B.

Hope it's clear.

Got confused with this step "x139n−12=x13 --> 9n=x259n=x25 -->" but got to know it's logic in the post mentioned at the bottom.

\(9n-12=x13\), so 9n - 12 equals to three-digit number x13, thus 9n = x13 + 12 = x25.

KarishmaB

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15 Feb 2021, 04:33

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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

\(q = 10^n – 49 = 99...99951\) (see the pattern 100 - 49 = 51; 1000 - 49 = 951; 10000 - 49 = 9951 and so on. As number of 0s increase, number of 9s keep increasing.)
Number of 9s will be (n - 2).

Sum of the digits is x13 i.e. 100x + 13.
We obtain 6 when we add 5 + 1 so all 9s should add up to 100x + 7.
If x = 1, we get 107 which is not a multiple of 9.
If x = 2, we get 207 which is a multiple of 9. So we must have 207/9 = 23 9s

Then n - 2 = 23 so n = 25.

Answer (B)

dthuy681

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24 Apr 2023, 10:55

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n=2 -> q=51
n=3 -> q=951
n=4 -> q=9951
--> Sum of digits of q = 9(n-2)+5+1 = 9(n-2)+6 = x13
--> 9(n-2) must be end up with number 7
-> (n-2) must end up with number 3
-> n must end up with number 5 --> 25 is correct

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26 Mar 2024, 01:55

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can i write the 3 digit number x13 as 100x + 10 + 3?

after which i wrote it as

10^{n} - 49 = 100x + 10 + 3
(n-2)9 - 5 + 1 = 100x + 10 + 3
(n-2)9 - 17 = 100x

then i plugged in values of N , started with B (& got lucky)

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