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Suppose n is an integer such that the sum of digits on n is 2, and 1010n. The number of different values of n is:
8
9
10
11
We have,
1010n and sum of digits for 'n' = 2
Clearly, nmin=10000000001
(1 followed by 9 zeros and finally 1)
Obviously, we can form 10 such numbers by shifting '1' by one place from right to left again and again.
Again, there is another possibility for 'n', n = 20000000000
So finally : No. of different values for n = 10 + 1 = 11
Hence, option 4 is the correct answer.
By: MIRZA SADDAM HUSSAIN ProfileResourcesReport error
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