1998 PMWC Problems/Problem I2

Problem I2

Triangular numbers and Square numbers can be represented in the following manner:

$[asy] int triangle(pair z, int n){ for(int i = 0; i < n; ++i){ for(int j = n-i; j > 0; --j){ dot((z.x+j -1 + i/2 ,z.y + i*sqrt(3)/2)); } } return 0; } triangle((0,0),2); label("3",(0.5,0),2S); triangle((5,0),3); label("6",(6,0),2S); triangle((11,0),4); label("10",(12.5,0),2S); int squ(pair z, int n){ for(int i = 0; i < n; ++i){ for(int j = 0; j < n; ++j){ dot((z.x + i , z.y + j)); } } return 0; } squ((0,-6),2); label("4",(0.5,-6),2S); squ((5,-6),3); label("9",(6,-6),2S); squ((11,-6),4); label("16",(12.5,-6),2S); //Credit to chezbgone2 for the diagram[/asy]$

Find a pair of consecutive Triangular Numbers and the difference between a pair of consecutive Square Numbers whose difference are both 11. What is the sum of these four numbers?

Solution

We want $n^2-(n-1)^2=11$ or $2n-1=11$ for $n=6$ . Therefore the two square numbers are $25$ and $36$ .

The $n$ th triangular number is equal to $n(n+1)/2$ . Therefore we want $\frac{n(n+1)}{2}=\frac{(n-1)n}{2}+11$ or $n(n+1)=n(n-1)+22$ . Solving gives $n=11$ so our two numbers are $55$ and $66$ .

The sum of all our numbers is therefore $25+36+55+66=\boxed{182}$ .

(Credit to 54math for the solution here)

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1998 PMWC Problems/Problem I2

Problem I2

Solution