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

Latest revision as of 19:48, 17 January 2024

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 $\boxed{\text{(C)}\ 120}$ .

Remark: The last number in each row ends in a perfect square because there is an odd amount of numbers in each row and the sum of the first n odd numbers is always a perfect square.

Solution 3 (eyeball it)

We can notice that even numbers are above even numbers so we can narrow our answers down to 2 solutions, $120$ and $122$ . Notice each number on the right end of each row is a perfect square. The perfect square closest to these is $121$ so $122$ is the first number on the next row. Notice $142$ is the third to last number on its row so it is too far away from 122, so our answer is $\boxed{\text{(C)}\ 120}$ .

@@ Line 16: / Line 16: @@
 Writing a couple more rows, the last number in each row ends in a perfect square. Thus <math>142</math> is two left from the last number in its row, <math>144</math>. One left and one up from <math>144</math> is the last number of its row, also a perfect square, and is <math>121</math>. This is one right and one up from <math>142</math>, so the number directly above <math>142</math> is one less than <math>121</math>, or <math>\boxed{\text{(C)}\ 120}</math>.
-==Solution 3==
+Remark: The last number in each row ends in a perfect square because there is an odd amount of numbers in each row and the sum of the first n odd numbers is always a perfect square.
-We can notice that even numbers are above even numbers so we can narrow our answers down to 2 solutions, <math>120</math> and <math>122</math>. Notice each number on the right end of each row is a perfect square. The perfect square closest to these is <math>121</math> so <math>122</math> is the first number on the next row. Notice <math>142</math> is the third to last number on its row so it is too far away from 122, so the answer is \fbox{C <math>120</math>}
+===Solution 3 (eyeball it)===
+We can notice that even numbers are above even numbers so we can narrow our answers down to 2 solutions, <math>120</math> and <math>122</math>. Notice each number on the right end of each row is a perfect square. The perfect square closest to these is <math>121</math> so <math>122</math> is the first number on the next row. Notice <math>142</math> is the third to last number on its row so it is too far away from 122, so our answer is <math>\boxed{\text{(C)}\ 120}</math>.
 ==See Also==
 {{AJHSME box|year=1993|num-b=23|num-a=25}}
 {{MAA Notice}}

1993 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
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