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

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What number is directly above <math>142</math> in this array of numbers?
 
What number is directly above <math>142</math> in this array of numbers?
  
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<cmath> \begin{tabular}{cccccc}& & & 1 & &\\ & & 2 & 3 & 4 &\\ & 5 & 6 & 7 & 8 & 9\\ 10 & 11 & 12 &\cdots & &\\ \end{tabular} </cmath>
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<cmath> \begin{array}{cccccc}& & & 1 & &\\ & & 2 & 3 & 4 &\\ & 5 & 6 & 7 & 8 & 9\\ 10 & 11 & 12 &\cdots & &\\ \end{array} </cmath>
  
 
<math>\text{(A)}\ 99 \qquad \text{(B)}\ 119 \qquad \text{(C)}\ 120 \qquad \text{(D)}\ 121 \qquad \text{(E)}\ 122</math>
 
<math>\text{(A)}\ 99 \qquad \text{(B)}\ 119 \qquad \text{(C)}\ 120 \qquad \text{(D)}\ 121 \qquad \text{(E)}\ 122</math>
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==Solution==
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===Solution 1===
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Notice that a number in row <math>k</math> is <math>2k</math> less than the number directly below it. For example, <math>5</math>, which is in row <math>3</math>, is <math>(2)(3)=6</math> less than the number below it, <math>11</math>.
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From row 1 to row <math>k</math>, there are <math>k \left(\frac{1+(-1+2k)}{2} \right) = k^2</math> numbers in those <math>k</math> rows. Because there are <math>12^2=144</math> numbers up to the 12th row, <math>142</math> is in the <math>k^{th}</math> row. The number directly above is in the 11th row, and is <math>22</math> less than <math>142</math>. Thus the number directly above <math>142</math> is <math>142-22=\boxed{\text{(C)}\ 120}</math>.
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===Solution 2===
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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>.
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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.
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===Solution 3 (eyeball it)===
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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>.
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==See Also==
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{{AJHSME box|year=1993|num-b=23|num-a=25}}
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{{MAA Notice}}

Latest revision as of 20:48, 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 $\boxed{\text{(C)}\ 120}$.

See Also

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
All AJHSME/AMC 8 Problems and Solutions

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