Some standardized exam, now long forgotton, included this following problem. The instructions were to circle the correct answer.
[#6] What whole number of two or more digits, if diminished by the sum of its digits, is divisible by exactly nine?
(a) There is no such number.
(b) The number eighty-four.
(c) Any even number.
(d) Any odd number.
(e) Any two digit number.
Task #1: Answer the question...
Task #2: Now, rethink...is it possible to answer this question correctly?
Hint: Try some numbers...a good choice, as suggested, is the number 84.
Solution Commentary: The problem becomes a bigger problem as you try to answer it.
First, the number 84 works, as [84-(8+4)]/9 = 8. So the answer is (b)?
Second, try some even numbers such as 24, 36, 48...they all work, so is the answer (c)?
Third, try some odd numbers such as 25, 37, 49...they all work, so is the answer (d)?
Note: It does not say "only even numbers" or "only odd numbers"...so maybe the answer is the collective (e).
Now try numbers such as 125, 236, 497...maybe the answer is ...wait, there is no option for three digit numbers specifically, and option (e) does not read "only two digit numbers."
What to do? You are taking this test and are being forced to circle one choice!
Relish the mathematical moment. Suppose you had the (m+1)-digit number nm...n1x, where m is one and greater. Then, the problem can be rewritten as [(10m+1nm+...+101+1n1+x)-(nm+...+n1+x)], which is a number divisible by 9. And, what happens if m=0...that is, are the mathematical aspects of the problem true for one-digit numbers as well?
So, the big quandary...What is the right answer...as only option (a) has been eliminated?