Not
a Correction, but An Additional Answer
Division M [December  Contest 2] Question 2B
ORIGINAL
ANSWER: 2000
NEWLY ACCEPTED
ALTERNATE ANSWER: 1.1
Credit
should be given to any student who had either 2000 or 1.1 as an
answer to this question.
PLEASE
GO BACK, MAKE THE CORRECTIONS BOTH ON THE PAPERS AND ON YOUR SCORING
AT OUR SECURE WEBSITE.
Our Division
M PICOs recently received a notification that they should accept
1.1 as an alternate answer to the above question, despite the
fact that, 2000 was listed as "the" correct answer on
the answer sheet. A big shout out goes to Jeremy Gold,
an eighth grader from The Rashi School, for coming up with a solution
that caught the entire Problem Writing Committee by surprise.
The problem
was…
A
palindrome reads the same forwards and backwards. The number
2017102 is a 7digit palindrome. Let A represent the least
palindrome greater than 2017102. Let B represent the greatest
palindrome less than 2017102. Find A  B. 
Jeremy's teacher,
Cindy Carter, submitted a very wellstated letter from
Jeremy regarding his way of looking at the problem. He felt that
the problem, as stated, did not disallow a rationalnumber palindrome
(it did not state that the palindromes, nor the difference between
them, had to be a whole number), so he decided to try what you
see below.
 Set A to
2017102.2017102 (the least rationalnumber palindrome that
is greater than 2017102);
 Set B to
2017101.1017102 (the greatest rationalnumber palindrome that
is less than 2017102).
 Subtracting
B from A, you get 1.1.
In all honesty,
when the Problem Writing Committee met over 10 months ago, we
never even entertained the notion of using a "rationalnumber
palindrome," but our directions for that problem allowed
Jeremy to think outside the box, and come up with a perfectly
acceptable answer to this question. Congratulations to Jeremy
for the creativity that we love for mathematicians to exhibit,
and thanks for setting this Problem Writing Committee straight!!
