N is a natural number of at least 5 digits and its leftmost digit is 6

Question

N is a natural number of at least 5 digits and its leftmost digit is 6. When this 6 is removed from N, the number thus obtained is found to be 1/25 times of N. What is the sum of the digits of N?

A. 9
B. 11
C. 13
D. 14
E. 18

jfranciscocuencag · Accepted Answer

6(10,000) + 1000x + 100y + 10z = N

N - 6(10,000) = \frac{N}{25}

N = \frac{N}{25} + 6(10,000)

25N = N+ 25*60,000

24N = 25*60,000

24N = \frac{25*60,000}{24}

N = 62,500

6 + 5 + 2 = 13

Apt0810 · Answer

Number N is 62500, since if 6 is removed then the number becomes 2500.

And then 2500*25= 62500.

Sum of the digits of N is 13

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chetan2u · Answer

Let the number be 6(X) where (X) is a n digit number..
So  6*10^n+X=25X....24X=6*10^n.....4X=10^n......X=100*10^{n-2}/4=25*10^{n-2} 
So we get the number X as 25000...
When we add the leftmost digit 6, we get 625000..
Sum =6+2+5+0+0+0...=13

C

aliakberza · Answer

Excellent question and some really good explanations here, so thanks everyone and esp  nick1816  for the question. Just wanted to add that while the methods described here are all great, I found using scientific notation made the cumbersome arithmetic just a little bit easier, allowing you to save those precious few seconds. Furthermore, using scientific notation helps us deal with the ambiguity with respect to the exact number of digits of N. So here it is:

Let's start with the simplest of cases where N has exactly 5 digits (at least 5 so it will have 5 or more):
 N=6BCDE  or  N=(6*10^4)+(B*10^3)+(C*10^2)+(D*10^1)+(E*10^0)

Now the question tells us that removing the 6, the leftmost digit, changes the number to make it  \frac{N}{25}

Algebraically, this can be expressed as:

N-(6*10^4)=\frac{N}{25}  --->  N-\frac{N}{25}=6*10^4  --->  \frac{24N}{25}=6*10^4  ---> N=6*10^4*\frac{25}{24} 
--->  N=6.25*10^4=62,500

Note that although we are unsure of whether N has 5 or more digits, we know that the sum of the digits will always be 13 and only the number of zeros after 625 will increase. By changing the leftmost digit in to a scientific notation we can see that the answer does not change no matter how many digits N may have.

The sum of the digits of N is then: 6+2+5=13, and the answer is C

Kinshook · Answer

Given: 
1. N is a natural number of at least 5 digits and its leftmost digit is 6. 
2. When this 6 is removed from N, the number thus obtained is found to be 1/25 times of N.

Asked: What is the sum of the digits of N?

Let the number be N=6x (digits denotion) where x is a natural number with at least 4 digits
 N= 6*10^{n} + x = 25 x  where n>=5
 6*10^{n} = 24x 
 x = \frac{6*10^{n}}{24} = \frac{10^n}{4} = \frac{100}{4} * 10^{n-2} = 25 * 10^{n-2}  
 N = 625 * 10^{n-2} 
Sum of digits = 6 + 2 + 5 = 13 since 0s do not contribute to sum of digits

IMO C

Mohammadmo · Answer

6*10000+1000a+100b+10c+d=
25000a+2500b+250c+25d
So,60000=24000a+2400b+240c+24d

2500=1000a+100b+10c+d
a=2,b=5,c&d=0
For number N; 625000
6+2+5=13
Option C

Posted from my mobile device

rvgmat12 · Answer

N/25 + 6*10^4 = N

N = 62500

N/25=2500

Sum of N's digits = 13

bumpbot · Answer

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N is a natural number of at least 5 digits and its leftmost digit is 6

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N is a natural number of at least 5 digits and its leftmost digit is 6

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