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

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===Solution 3===
 
===Solution 3===
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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 the answer is <math>\boxed{\text{(C)}\120}</math>.
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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>.
  
 
==See Also==
 
==See Also==
 
{{AJHSME box|year=1993|num-b=23|num-a=25}}
 
{{AJHSME box|year=1993|num-b=23|num-a=25}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 18:42, 19 October 2021

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


Solution 3

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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