The Digit Dilemma
There is a number with the property that when its digits are added together, it gives a result which is one-third the value of the number itself. What is the number?
The answer is 27.
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1 solution
Moderator note:
You should first present the argument that we must have a two digit number. Alternatively, you could say "first we will consider the case where we have a two digit number".
What is the minimum value of , where are digits?
The problem says "digits" so the number is at the least a 2 digit number. Let the digits of the number be "x" and "y". Hence the number is- xy. Expanding xy I can rewrite it as 10x+y (eg- 45 can be written as 10*4+5) Hence, 1/3(10x+y)=(x+y) or 10x+y= 3(x+y) or 10x+y=3x+3y or 7x=2y Since 7 and 2 are prime numbers the only numbers we can multiply them with to get the same result would be themselves. Hence, x=2 and y=7 So the number is 27.
Note: Why have I assumed the number to be a two digit number?
Let the number not be a two digit number. Hence it could be a three digit number or higher. Consider it to be a three digit number first. So say the digits are x,y and z. Hence the number is xyz or 100x+10y+z. Now according to the question the sum of the three digits multiplied by 3 would give us 3x+3y+3z. Now if you notice 100x+10y+z is too large to be equal to 3x+3y+3z. Why? 1) 10y> 3y 2) 100x+z> 3x+3z How is 100x+z> 3x+3z? The minimum value for x and the maximum value for z here are 1 and 9 respectively. Using them in the equation we would have 100x+z as 109 and 3x+3z as 30. One could argue why use 1 and 9 because that is the only way to maximize the RHS and minimize the LHS; since there is a 100 in the co-efficient of x. This proves that for a three digit number the question is not applicable as the value of the number would be many times larger than the sum of its digits. Hence, this subsequently proves that greater numbers(4 digit, 5 digit etc) will be extremely large for the statement to stand true.