Difference between revisions of "2004 AMC 12B Problems/Problem 22"
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==Solution B== | ==Solution B== | ||
− | All the unknown entries can be expressed in terms of b. | + | All the unknown entries can be expressed in terms of <math>b</math>. |
− | Since 100e = beh = ceg = def, it follows that h = 100 | + | Since <math>100e = beh = ceg = def</math>, it follows that <math>h = \frac{100}{b}, g = \frac{100}{c}</math>, |
− | and f = 100/ | + | and <math>f = \frac{100}{d}</math>. Comparing rows <math>1</math> and <math>3</math> then gives |
− | 50bc = 2 | + | <math>50bc = 2 \cdot \frac{100}{b} \cdot \frac{100}{c}</math>, |
− | from which c = 20/ | + | from which <math>c = \frac{20}{b}</math>. |
− | Comparing columns 1 and 3 gives | + | Comparing columns <math>1</math> and <math>3</math> gives |
− | 50d | + | <math>50d \cdot \frac{100}{c}= 2c \cdot \frac{100}{d}</math>, |
− | from which d = c | + | from which <math>d = \frac{c}{5} = \frac{4}{b}</math>. |
− | Finally, f = 25b, g = 5b, and e = 10. All the entries are positive integers | + | Finally, <math>f = 25b, g = 5b</math>, and <math>e = 10</math>. All the entries are positive integers |
− | if and only if b = 1, 2, or 4. The corresponding values for g are 5, 10, and | + | if and only if <math>b = 1, 2,</math> or <math>4</math>. The corresponding values for <math>g</math> are <math>5, 10,</math> and |
− | 20, and their sum is <math>\boxed{\mathbf{(C)}35} </math>. | + | <math>20</math>, and their sum is <math>\boxed{\mathbf{(C)}35} </math>. |
Credit to Solution B goes to http://billingswest.billings.k12.mt.us/math/AMC%201012/AMC%2012%20work%20sheets/2004%20AMC%2012B%20ws-15.pdf, a page with a play-by-play explanation of the solutions to this test's problems. | Credit to Solution B goes to http://billingswest.billings.k12.mt.us/math/AMC%201012/AMC%2012%20work%20sheets/2004%20AMC%2012B%20ws-15.pdf, a page with a play-by-play explanation of the solutions to this test's problems. |
Revision as of 10:12, 20 August 2017
Contents
Problem
The square
is a multiplicative magic square. That is, the product of the numbers in each row, column, and diagonal is the same. If all the entries are positive integers, what is the sum of the possible values of ?
Solution A
If the power of a prime other than divides , then from it follows that , but then considering the product of the diagonals, but , contradiction. So the only prime factors of are and .
It suffices now to consider the two magic squares comprised of the powers of and of the corresponding terms. These satisfy the normal requirement that the sums of rows, columns, and diagonals are the same, owing to our rules of exponents; additionally, all terms are non-negative.
The powers of :
So , so . Indeed, we have the magic squares
The powers of :
Again, we get . However, if we let , then , which obviously gives us a contradiction, and similarly for . For , we get
In conclusion, can be , and their sum is .
Solution B
All the unknown entries can be expressed in terms of . Since , it follows that , and . Comparing rows and then gives , from which . Comparing columns and gives , from which . Finally, , and . All the entries are positive integers if and only if or . The corresponding values for are and , and their sum is .
Credit to Solution B goes to http://billingswest.billings.k12.mt.us/math/AMC%201012/AMC%2012%20work%20sheets/2004%20AMC%2012B%20ws-15.pdf, a page with a play-by-play explanation of the solutions to this test's problems.
See also
2004 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 21 |
Followed by Problem 23 |
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 AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.