SOLUTION: How many 2-digit numbers are there in which the sum of its digits drop to its previous half when you add a four to it.

Question 1066161: How many 2-digit numbers are there in which the sum of its digits drop to its previous half when you add a four to it.
Found 2 solutions by ikleyn, MathTherapy:
Answer by ikleyn(52905)   (Show Source): You can put this solution on YOUR website!
.
What is "its previous half" ???

Who is the author/the source of this post ???

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Comment from student: Sorry, it's not "previous half" it's actually just "its half"
This is the full question; How many 2-digit numbers are there in which the sum of its digits drop to its half when you add a four to it.
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My response: Slightly better, but still is not clear.

When do you add 4 to it: before calculating the sum of the digits or after it???

I do not believe that such dirty formulation can come from the textbook or from the teacher assignment.

Again: Who is the author/the source of this post ???

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Comment from student: How many 2-digit numbers are there in which the sum of its digits drop to its half when you add a four to the number.
Found it in a math olympiad book
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My diagnosis. Fantastically dirty formulation for the math problem and disgrace for the book, for its author, for the editor and/or for the translator.

My advice: forget about this problem.

    WHO   IS   THE   AUTHOR?

Answer by MathTherapy(10557)   (Show Source): You can put this solution on YOUR website!

How many 2-digit numbers are there in which the sum of its digits drop to its previous half when you add a four to it.

If I understand this right, is this saying that when 4 is added to the number, the sum of its digits becomes of the sum of the digits of the original number?
If so, then your number is: , since the sum of its digits is 10 (2 + 8), and when 4 is added to the number, the result is 32 (28 + 4), whose digits
sum to 5 (3 + 2), and which is actually of 2 + 8, or of 10.
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