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?

q is of the from 99999....51

sum of digits will be 9a+5+1 =9a+6

so 9a +6 = x13

or 9a =x07

x is 2 as only 207 is divisible by 9

so 207/9 = 23

i.e 23 9's are reqd + 51 i.e total 25 digits

so reqd nos should have 25 0's i.e n= 25

B

Woh! What an approach. Really appreciable ...

10^n-49=10...0-49. The 10...0 with n zeros and n+1 digits. So 10...0 - 49 = 9..951 will have n digits and n-2 nines. The summ is 9*(n-2)+5+1=x13 <=> 9*n-18+6=x13 <=> 9*n - 12=x13 <=> 9*n=x25. You can only obtain a five in the unit digit of the product 9*n if the unit digit of n is 5 (since 9*5=45). So the answer is n=25 (the only one with a five in the unit digit).

we can see that 10^2 has 2 zeroes

10^3 has 3 zeroes and so on ...

so go by option starting with E if there 28 n den 10^28 will have 28 zeroes and if u subtract 49 from it u will have 9999....51 last two digit will have sum 6 always and there will be 3 less number of nine from 10^n value so there will be 25 9 and its sum is 25*9=225 and add 6 to it from 51 so it becumes 231 but we have been given its sum is x13 as soon as u dont get the last digit same you go to next option and since the option are in ascending order from A to E u simply have to subtract 9 from prev sum so u hv ans:

gabriel wrote:The question can be found on www.manhattangmat.com

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

Regards