November "Fun with Math" answers
Lots of entrants figured out palindrome
answers
Young Saint Louis.com attracted some first-time
entries for the November "Fun with Math" questions.
And there were 16 individual and one class-wide winners.
FM creator George Yu was a little concerned when
he proposed a series of questions involving palindrome numbers.
He needn't have worried since YSL.com entrants caught
on very well.
One of the new entries for "Fun with Math" was from
teacher Amy Goldman's 6th grade class at Parkway West Middle
School in Chesterfield. She sent in a single entry for the
whole class and all the answers were correct.
Other first-time entries came from students in teacher Steven
Ondes' class at Liberty Middle School in Edwardsville, IL.
Those students also understood palindromes well.
As last month, when we had multiple winners, YSL.com
chose to award three $10 Borders gift certificates, instead
of the usual two. The three were chosen randomly from the
total number of all-correct entries.
The three $10 Borders gift certificate awardees were:
Samay Gupchup, Edwardsville, IL; Durga Kullakanda,
Maryland Heights, MO, and Chad Turner, Glen Carbon,
IL.
The other 13 individual entrants with all correct entries
were:
John Bentley, Edwardsville, IL; Shaun Brumgard,
Glen Carbon, IL; Monica Chavan, Edwardsville, IL; Quenton
Crane, Edwardsville, IL; Will Jeziorshi, Glen Carbon,
IL; Kim Johnson, Edwardsville, IL; Dani Kirsch,
Edwardsville, IL; Emily Lange, Edwardsville, IL; Natasha
Meinzen, Edwardsville, IL: Henry Lu, Edwardsville,
IL; Jacob Veitch, Edwardsville, IL; Ryan Wahidi,
Creve Coeur, MO, and Ryan Zhong, Edwardsville, IL.
(Editor's note: YSL.com has made a sincere effort
to get the spelling of winners names correct. But, frankly,
some of your printing is very difficult to read. Please take
care in printing your names when making future entries.)
(If you'd like to enter the December "Fun with Math"
contest, just click here.)
Solution for November's
Fun with Math
-
Numbers greater than 99
and less than 200 have exactly three digits. Also, these
numbers must have a 1 in the hundred's place. It must
also have a 1 in the one's place because it is a palindrome.
That leaves only the ten's place. The ten's place could
be any digit from 0 to 9. That results in 10 palindromes:
101, 111, 121, 131, 141, 151, 161, 171, 181, 191.
-
The range of numbers given
is exactly all the three-digit numbers. The ten's place
can be any digit from 0 to 9. The one's and hundred's
place are the same because the number is a palindrome.
Those two places can be any digit from 1 to 9. If the
hundred's place was a 0, then the number would not be
three digits. The ten's place can be ten digits and the
one's and hundred's place can be nine digits.
9 (10) = 90
101, 111, 121, 131, 141, 151, 161, 171, 181, 191 202,
212, 222, 232, 242, 252, 262, 272, 282, 292 303, 313,
323, 333, 343, 353, 363, 373, 383, 393 404, 414, 424,
434, 444, 454, 464, 474, 484, 494 505, 515, 525, 535,
545, 555, 565, 575, 585, 595 606, 616, 626, 636, 646,
656, 666, 676, 686, 696 707, 717, 727, 737, 747, 757,
767, 777, 787, 797 808, 818, 828, 838, 848, 858, 868,
878, 888, 898 909, 919, 929, 939, 949, 959, 969, 979,
989, 999
-
The next palindrome will
have different digits from 46964. We can either change
the 4's, the 6's, the 9, or a combination of them. If
we change the 4's, we will increase the ten-thousand's
place. If we change the 6's, we will increase the thousand's
place. If we change the 9, we will only change the hundred's
place.
Since changing the 9 would result in the smallest increase,
we want to change the 9 and no other digits. However,
changing only the 9 (to a digit 0-8) makes the number
smaller. We have to change the 6's.
After changing the 6's to 7's, we have 47974. Now, we
still don't want to change the 4's. But we can make 47974
smaller by changing the 9 to a 0. The next palindrome
after 46964 is 47074.
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