Difference between revisions of "1989 AIME Problems/Problem 1"
m (→Solution 2) |
Talkinaway (talk | contribs) (→Solution) |
||
| Line 8: | Line 8: | ||
=== Solution 2=== | === Solution 2=== | ||
Note that the four numbers to multiply are symmetric with the center at <math>29.5</math>. Multiply the symmetric pairs to get <math>31\cdot 28=868</math> and <math>30\cdot 29=870</math>. Now clearly <math>868\cdot 870 + 1 = (869-1)(869+1) + 1 = 869^2 - 1^2 + 1 = 869^2</math>. | Note that the four numbers to multiply are symmetric with the center at <math>29.5</math>. Multiply the symmetric pairs to get <math>31\cdot 28=868</math> and <math>30\cdot 29=870</math>. Now clearly <math>868\cdot 870 + 1 = (869-1)(869+1) + 1 = 869^2 - 1^2 + 1 = 869^2</math>. | ||
| + | |||
| + | === Solution 3=== | ||
| + | The last digit under the radical is <math>1</math>, so the square root must either end in <math>1</math> or <math>9</math>, since <math>x^2 = 1\pmod {10}</math> means <math>x = \pm 1</math>. Additionally, the number must be near <math>29 \cdot 30 = 870</math>, narrowing the reasonable choices to <math>869</math> and <math>871</math>. | ||
| + | |||
| + | Continuing the logic, the next-to-last digit under the radical is the same as the last digit of <math>28 \cdot 29 \cdot 3 \cdot 31</math>, which is <math>6</math>. Quick computation shows that <math>869^2</math> ends in <math>61</math>, while <math>871^2</math> ends in <math>41</math>. Thus, the answer is <math>\boxed{869}</math>. | ||
| + | |||
| + | ===Solution 4=== | ||
| + | Similar to Solution 1 above, call the consecutive integers <math>(n-\frac{3}{2}), (n-\frac{1}{2}), (n+\frac{1}{2}), (n+\frac{3}{2})</math> to make use of symmetry. Note that <math>n</math> itself is not an integer - in this case, <math>n = 29.5</math>. The expression becomes <math>\sqrt{(n-\frac{3}{2})(n + \frac{3}{2})(n - \frac{1}{2})(n + \frac{1}{2}) + 1}</math>. Distributing each pair of difference of squares first, and then distributing the two resulting quadratics and adding the constant, gives <math>\sqrt{n^4 - \frac{5}{2}n^2 + \frac{25}{16}}</math>. The inside is a perfect square trinomial, since <math>b^2 = 4ac</math>. It's equal to <math>\sqrt{(n^2 - \frac{5}{4})^2}</math>, which simplifies to <math>n^2 - \frac{5}{4}</math>. You can plug in the value of <math>n</math> from there, or further simplify to <math>(n - \frac{1}{2})(n + \frac{1}{2} - 1)</math>, which is easier to compute. Either way, plugging in <math>n=29.5</math> gives <math>\boxed{869}</math>. | ||
== See also == | == See also == | ||
Revision as of 08:59, 21 August 2012
Contents
Problem
Compute 
.
Solution
Solution 1
Let's call our four consecutive integers 
. Notice that 
. Thus, 
.
Solution 2
Note that the four numbers to multiply are symmetric with the center at 
. Multiply the symmetric pairs to get 
and 
. Now clearly 
.
Solution 3
The last digit under the radical is 
, so the square root must either end in or 
, since 
means 
. Additionally, the number must be near 
, narrowing the reasonable choices to 
and 
.
Continuing the logic, the next-to-last digit under the radical is the same as the last digit of 
, which is 
. Quick computation shows that 
ends in 
, while 
ends in 
. Thus, the answer is 
.
Solution 4
Similar to Solution 1 above, call the consecutive integers 
to make use of symmetry. Note that 
itself is not an integer - in this case, 
. The expression becomes 
. Distributing each pair of difference of squares first, and then distributing the two resulting quadratics and adding the constant, gives 
. The inside is a perfect square trinomial, since 
. It's equal to 
, which simplifies to 
. You can plug in the value of from there, or further simplify to 
, which is easier to compute. Either way, plugging in 
gives .
See also
| 1989 AIME (Problems • Answer Key • Resources) | ||
| Preceded by First Question |
Followed by Problem 2 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
