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?
  
\[\begin{tabular}{cccccc}
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<cmath> \begin{array}{cccccc}& & & 1 & &\\ & & 2 & 3 & 4 &\\ & 5 & 6 & 7 & 8 & 9\\ 10 & 11 & 12 &\cdots & &\\ \end{array} </cmath>
& & & 1 & & \\
 
& & 2 & 3 & 4 & \\
 
& 5 & 6 & 7 & 8 & 9 \\
 
10 & 11 & 12 & \cdots & & \\
 
\end{tabular}\]
 
  
 
<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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==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 18:55, 10 March 2015

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

See Also

1993 AJHSME (ProblemsAnswer KeyResources)
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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