Quote:
Originally Posted by Frosty
My wife volunteers at the middle school to coach kids for a local math contest called Math Is Cool. For the contest, kids have to take a variety of tests, some individual and some team tests, as well as a "College Bowl" type team to team competition.
Anyway, to prepare for teaching the kids, she works through the problems on old tests so she can explain to the kids at practice. She ran into this problem and couldn't figure out a quick solution:
Now you can brute force the answer by listing out all 90 numbers but these are timed tests and that would take too long. She did it that way to see if there was a pattern but we couldn't see one that could she could explain.
First of all, you have the nine "10's": 10, 20, 30, 40, etc
Then you have the nine "9's": 18, 27, 36, 45, etc
After that, there are only six other numbers:
two "3's": 12, 21, two "6's": 24, 42, and and two "12's: 48, 84
(when I say a number is a "9's", I mean the digits add up to 9: 18 > 1+8=9)
That adds up to 24. However, 90 is on both the 10's and 9's list so you have to subtract one to get 23.
We also noticed that all multiple of 12 and of 21 are on the list but are not sure how that helps.
So - is there a quick way to do this problem to explain to the kids if they run into something like this?
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You're doing it wrong.
Go through the sums, starting with zero.
0, none
1, 1+0, none
2, 1+1 and 2+0, bam 20
3, 1+2, 2+1, 3+0 bam 12, 21, and 30
4, 1+3, 3+1, 2+2, 4+0, bam 40
that is going to be quicker than looking at all 90. For instance 5 you can see in about an instant that 50 is the only one. Its reasonable to do the calculations in a minute or so if you have the concept down.