1998 AJHSME Problems/Problem 24
Problem
A rectangular board of 8 columns has squared numbered beginning in the upper left corner and moving left to right so row one is numbered 1 through 8, row two is 9 through 16, and so on. A student shades square 1, then skips one square and shades square 3, skips two squares and shades square 6, skips 3 squares and shades square 10, and continues in this way until there is at least one shaded square in each column. What is the number of the shaded square that first achieves this result?
![[asy] unitsize(20); for(int a = 0; a < 10; ++a) { draw((0,a)--(8,a)); } for (int b = 0; b < 9; ++b) { draw((b,0)--(b,9)); } draw((0,0)--(0,-.5)); draw((1,0)--(1,-1.5)); draw((.5,-1)--(1.5,-1)); draw((2,0)--(2,-.5)); draw((4,0)--(4,-.5)); draw((5,0)--(5,-1.5)); draw((4.5,-1)--(5.5,-1)); draw((6,0)--(6,-.5)); draw((8,0)--(8,-.5)); fill((0,8)--(1,8)--(1,9)--(0,9)--cycle,black); fill((2,8)--(3,8)--(3,9)--(2,9)--cycle,black); fill((5,8)--(6,8)--(6,9)--(5,9)--cycle,black); fill((1,7)--(2,7)--(2,8)--(1,8)--cycle,black); fill((6,7)--(7,7)--(7,8)--(6,8)--cycle,black); label("$2$",(1.5,8.2),N); label("$4$",(3.5,8.2),N); label("$5$",(4.5,8.2),N); label("$7$",(6.5,8.2),N); label("$8$",(7.5,8.2),N); label("$9$",(0.5,7.2),N); label("$11$",(2.5,7.2),N); label("$12$",(3.5,7.2),N); label("$13$",(4.5,7.2),N); label("$14$",(5.5,7.2),N); label("$16$",(7.5,7.2),N); [/asy]](http://latex.artofproblemsolving.com/b/d/a/bdae457892201d1de7abac9931026ed0fce5dc70.png)
Solution
Solution 1
The numbers that are shaded are the triangular numbers, which are numbers in the form 
for positive integers. They can also be generated by starting with 
, and adding 
as in the description of the problem.
Squares that have the same remainder after being divided by 
will be in the same column. Thus, we want to find when the last remainder, from 
to 
, is found.
So, instead of adding 
, we can effectively either add 
or subtract 
if we are only concerned about remainders when divided by . We will pick the number that keeps the terms on the list between and . We get:
Finally, a term with is found, and checking, all numbers through are also on the right side of the list. This means the last term in our sequence is the first time that column is shaded. There are 
terms in the sequence, leading to an answer of 
, which is choice 
.
Solution 2
Note that the triangular numbers up to 
are 
. When you divide each of those numbers by , all remainders must be present. We first search for number(s) that are evenly divisible by ; if two such numbers exist, we search for numbers that leave a remainder of , etc.
Quickly scanning the list, only 
and are even. That smaller list doesn't have any multiples of until it hits . So 
must be the answer.
Solution 3
The numbers shaded are triangular numbers of the form . For this number to be divisible by , the numerator must be divisible by 
. Since only one of 
and 
can be even, only one of them can have factors of 
. Therefore, the first time the whole expression is divisible by is when either 
or when 
. This gives 
as the first time 
is divisible by , which gives . No other triangular number lower than that is divisible by , and thus the 
column on the checkerboard won't be filled until then. That gives as the right answer.
See also
| 1998 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 23 |
Followed by Problem 25 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
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