Difference between revisions of "1993 AJHSME Problems/Problem 24"
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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>. | 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>. | ||
+ | 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=== | + | ===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>. | 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>. | ||
Latest revision as of 19:48, 17 January 2024
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 .
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, and . Notice each number on the right end of each row is a perfect square. The perfect square closest to these is so is the first number on the next row. Notice is the third to last number on its row so it is too far away from 122, so our answer is .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.