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8th Graders olimpyad problem


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From 9 to 90.

910111213...888990 divided by 990

= 910111213...88899 divided by 99

It's divisible by 9 if all digits sum to 9. The digits 1-9 sum to 45, and 4+5=9, so the "ones" digits of each set of numbers like 10-19 or 30-39 would be divisible by 9. As for the "tens" digits, there are ten numbers each in the range from 10-19, 20-29, 30-39, ... 80-89, (plus the starting 9 and ending 90 which wouldn't affect divisibility by 9). Pair up the tens digits: 10-19 & 80-89, 20-29 & 70-79 ... 40-49 & 50-59, and it's obvious that the tens digits will also be divisible by 9. So that whole number is divisible by 9.

You also need to know that a number is divisible by 11 if and only if taking alternating sums and differences of its digits leads to a number that's divisible by 11. In this case that would be

9 - 1 + 0 - 1 + 1 - 1 + 2 - 1 + 3 .... - 8 + 8 - 8 + 9 - 9 + 0

It should be obvious that if you consider all the numbers from 10 to 89, there are 10 each of digits 1-8 in the tens spot (which will be added) and 8 each of the digits 1-9 in the ones spot (which will be subtracted), and the leading 9 and trailing 90 will cancel each other out. The digits in the tens spot will sum to 10x(1+2+...+7+8) = 10x(9x4) = 360, and the digits in the ones spot will sum to 8x(1+2+...+8+9) = 8x45 = 360. So it's divisible by 11.

Edit: Out of curiosity, what sort of time and resources do the kids get to solve these problems?

Edited by plasmid
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From 9 to 90.

910111213...888990 divided by 990

= 910111213...88899 divided by 99

It's divisible by 9 if all digits sum to 9. The digits 1-9 sum to 45, and 4+5=9, so the "ones" digits of each set of numbers like 10-19 or 30-39 would be divisible by 9. As for the "tens" digits, there are ten numbers each in the range from 10-19, 20-29, 30-39, ... 80-89, (plus the starting 9 and ending 90 which wouldn't affect divisibility by 9). Pair up the tens digits: 10-19 & 80-89, 20-29 & 70-79 ... 40-49 & 50-59, and it's obvious that the tens digits will also be divisible by 9. So that whole number is divisible by 9.

You also need to know that a number is divisible by 11 if and only if taking alternating sums and differences of its digits leads to a number that's divisible by 11. In this case that would be

9 - 1 + 0 - 1 + 1 - 1 + 2 - 1 + 3 .... - 8 + 8 - 8 + 9 - 9 + 0

It should be obvious that if you consider all the numbers from 10 to 89, there are 10 each of digits 1-8 in the tens spot (which will be added) and 8 each of the digits 1-9 in the ones spot (which will be subtracted), and the leading 9 and trailing 90 will cancel each other out. The digits in the tens spot will sum to 10x(1+2+...+7+8) = 10x(9x4) = 360, and the digits in the ones spot will sum to 8x(1+2+...+8+9) = 8x45 = 360. So it's divisible by 11.

Edit: Out of curiosity, what sort of time and resources do the kids get to solve these problems?

 

 

Usually 45 minutes per problem with pen, paper, ruler, compasses. Rarely they accept calculater and/or formula book.

 

But state olimpyad is 90 minutes per problem, 3 problems, 2 day. I once sat 9 hours straight coz I was challengin in above grade too.

Edited by Barcallica
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I am adding this using my mobile and I don't see any spoiler button, so the mod may please put this in a spoiler.
Here is a general method to solve for more possibilities

 

Let ab = x
Then cd = 99 -x
Abcd = 100x + 99 -x = 99 (x+1)
Now the condition is that N (x,99-x) should be divisible by 99 (x+1)

All N are divisible by 9 and 11 - easy to prove.
Now we just need to see which x+1 can divide N.
Some numbers I can identify quickly are
1,98 divisible by 2
4,95 divisible by 5
7,92 divisible by 8
9,90 divisible by 10
19,80 divisible by 20
24,75 divisible by 25
39,60 divisible by 40
49,50 divisible by 50

There can be more possibilities. For example I don't remember the divisibility rules for 7,13, 27, 17, 19, 23 etc.

So can not check for multiples of 3 or above numbers

Edited by bonanova
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