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? | ||
− | + | <cmath> \begin{array}{cccccc}& & & 1 & &\\ & & 2 & 3 & 4 &\\ & 5 & 6 & 7 & 8 & 9\\ 10 & 11 & 12 &\cdots & &\\ \end{array} </cmath> | |
− | |||
− | & & 2 & 3 & 4 & \\ | ||
− | & 5 & 6 & 7 & 8 & 9 \\ | ||
− | 10 & 11 & 12 & \cdots & & \\ | ||
− | \end{ | ||
<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> | ||
+ | |||
+ | ==Solution== | ||
+ | ===Solution 1=== | ||
+ | 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>. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | ===Solution 2=== | ||
+ | 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>. | ||
+ | |||
+ | ==See Also== | ||
+ | {{AJHSME box|year=1993|num-b=23|num-a=25}} | ||
+ | {{MAA Notice}} |
Latest revision as of 18:55, 10 March 2015
Problem
What number is directly above in this array of numbers?
Solution
Solution 1
Notice that a number in row is less than the number directly below it. For example, , which is in row , is less than the number below it, .
From row 1 to row , there are numbers in those rows. Because there are numbers up to the 12th row, is in the row. The number directly above is in the 11th row, and is less than . Thus the number directly above is .
Solution 2
Writing a couple more rows, the last number in each row ends in a perfect square. Thus is two left from the last number in its row, . One left and one up from is the last number of its row, also a perfect square, and is . This is one right and one up from , so the number directly above is one less than , or .
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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