When 10^{60} – m, where m is a two-digit positive integer, written in

Question

When  10^{60} – m , where  m  is a two-digit positive integer, written in the decimal notation, the sum of the digits of the resulting number is 533. How many distinct values of  m  exist?

(A)5

(B) 6

(C) 7

(D) 8

(E) 9

Gaurav2896 · Answer

Hi  sjuniv32  is the answer C? Can you confirm once , i got D ->
sum in 10^60-m = 9*58 + (a+b)
=>522 + a+b
so a+b = 11.
So a and b can be (2,9), (3,8),(4,7),(5,6),(6,5),(7,4),(8,3),(9,2). For all these 8 terms m will be distinct .So answer should be 8.
IMO D. Help if i have done some mistake.

Krutak · Answer

(9,2) will not be possible as 2 digit number is being subtracted. 
m = two digit number

Posted from my mobile device

Bogei · Answer

Hi there, still fairly new posting reply so please excuse the format:

Our goal is to find all m's such that sum of the digits of  10^{60} - m = 533

For a simpler case: Let's assume  m = 99   and  10^4  instead of  10^{60} ;  10^4- m  we would have  10000-99 = 9901 ,

From that we can see for every n in  10^n , we will have at least  n-2  9's (in our case  n = 4   and number of 9's is  n-2 )

so for  10^{60}   we would have 58 9's and the sum of digits so far would be   (58 * 9) = 522,  we are still missing  533-522 = 11.

Keep in mind we are  not  looking for all m's such that the digits of m add up to 11, we are looking for all m's such that the digits of  100-m  is 11. (Observe that I'm using 100 here since we are only concerned with the last 2 digits of the sum that add up to 11, the rest are all 9's as stated above)

so we need the sum of digits of  100-m  equal to 11, m could be

m = 17 -> 100 - 17 = 83, 8+3=11 
 m = 26 -> 100 - 26 = 74, 7+4 =11 , and so on. Also observe the reverse of the digits also works (e,g 71, 62)

in the end we have 7 cases (17, 26, 35, 44, 53, 62, 71). Answer is C

bumpbot · Answer

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When 10^{60} – m, where m is a two-digit positive integer, written in

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When 10^{60} – m, where m is a two-digit positive integer, written in

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