# Difference between revisions of "1993 AJHSME Problems/Problem 24"

## Problem

What number is directly above $142$ in this array of numbers?

$$\begin{array}{cccccc}& & & 1 & &\\ & & 2 & 3 & 4 &\\ & 5 & 6 & 7 & 8 & 9\\ 10 & 11 & 12 &\cdots & &\\ \end{array}$$

$\text{(A)}\ 99 \qquad \text{(B)}\ 119 \qquad \text{(C)}\ 120 \qquad \text{(D)}\ 121 \qquad \text{(E)}\ 122$

## Solution

### Solution 1

Notice that a number in row $k$ is $2k$ less than the number directly below it. For example, $5$, which is in row $3$, is $(2)(3)=6$ less than the number below it, $11$.

From row 1 to row $k$, there are $k \left(\frac{1+(-1+2k)}{2} \right) = k^2$ numbers in those $k$ rows. Because there are $12^2=144$ numbers up to the 12th row, $142$ is in the $k^{th}$ row. The number directly above is in the 11th row, and is $22$ less than $142$. Thus the number directly above $142$ is $142-22=\boxed{\text{(C)}\ 120}$.

### Solution 2

Writing a couple more rows, the last number in each row ends in a perfect square. Thus $142$ is two left from the last number in its row, $144$. One left and one up from $144$ is the last number of its row, also a perfect square, and is $121$. This is one right and one up from $142$, so the number directly above $142$ is one less than $121$, or $\boxed{\text{(C)}\ 120}$.