If x+y=10^15, where x and y are positive integers, y=? (1) The sum of

Question

If x+y=10^15, where x and y are positive integers, y=?

(1) The sum of all digits of x is 130.
(2) y is greater than 50 and less than 100.

IanStewart · Answer

10^15 is the smallest possible sixteen-digit number. If x and y are positive, and x+y = 10^15, then x can have at most 15 digits.

If the sum of all of the digits of x is 130, notice that x must have exactly fifteen digits; if it had only 14, the greatest digit sum we could have would be 9*14 = 126. But if x has fifteen digits, there are several ways to get a digit sum of 130: x can be, say, 499999999999999 or 99999999999994, for example. So Statement 1 is not sufficient. Statement 2 is also clearly insufficient alone.

If we use both Statements, then x must be just less than 10^15, between 50 and 100 less. So the last three digits of x will be strictly between 900 and 950. Notice then that the first thirteen digits of x must all be 9's, so they will add up to 117. The last two digits, which are strictly between 00 and 50, so are between 01 and 49 inclusive, must therefore add to 13 to get the required sum of 130. But if the tens digit is 4 or smaller, the only way to get a sum of 13 is if the last two digits are precisely 49. So the answer is C, and x must be equal to 99999999999949.

chetan2u · Answer

x+y=10^{15} 
(1) The sum of all digits of x is 130.
number of digits in  10^{15}  is 15+1=16, that is digit 1 followed by 15 zeroes.
If the sum of digits of x is 130, the minimum number of digits is next integer after  \frac{130}{9}  or  14\frac{4}{9} , so number of digits is 15.
The largest value of x = 999,999,999,999,994, and y = 6
the next value of x will be x = 999,999,999,999,949, and y = 51, the next value of y = 501 and so on
insuff

(2) y is greater than 50 and less than 100.
 y could be anything between 50 and 100.

Combined
y can be 6, 51, 501, 5001,......
y is between 50 and 100.
Thus y is 51.

C

If x+y=10^15, where x and y are positive integers, y=? (1) The sum of

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