Difference between revisions of "1989 AIME Problems/Problem 1"
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== Solution 3 (Symmetry with Generalization) == | == Solution 3 (Symmetry with Generalization) == | ||
− | + | More generally, we can prove that one more than the product of four consecutive integers must be a perfect square: | |
− | |||
− | |||
− | More generally, we can | ||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
− | (a+3)(a+2)(a+1)(a)+1 &= | + | (a+3)(a+2)(a+1)(a)+1 &= [(a+3)(a)][(a+2)(a+1)]+1 \\ |
− | &= | + | &= [a^2+3a][a^2+3a+2]+1 \\ |
− | &= | + | &= [a^2+3a]^2+2[a^2+3a]+1 \\ |
− | + | &= [a^2+3a+1]^2. | |
− | &= | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
At <math>a=28,</math> we have <cmath>\sqrt{(a+3)(a+2)(a+1)(a)+1}=a^2+3a+1=\boxed{869}.</cmath> | At <math>a=28,</math> we have <cmath>\sqrt{(a+3)(a+2)(a+1)(a)+1}=a^2+3a+1=\boxed{869}.</cmath> | ||
− | |||
~Novus677 (Fundamental Logic) | ~Novus677 (Fundamental Logic) | ||
~MRENTHUSIASM (Reconstruction) | ~MRENTHUSIASM (Reconstruction) | ||
− | == Solution | + | == Solution 4 (Symmetry with Generalization) == |
− | + | Similar to Solution 1 above, call the consecutive integers <math>\left(n-\frac{3}{2}\right), \left(n-\frac{1}{2}\right), \left(n+\frac{1}{2}\right), \left(n+\frac{3}{2}\right)</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{\left(n-\frac{3}{2}\right)\left(n + \frac{3}{2}\right)\left(n - \frac{1}{2}\right)\left(n + \frac{1}{2}\right) + 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{\left(n^2 - \frac{5}{4}\right)^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>\left(n - \frac{1}{2}\right)\left(n + \frac{1}{2}\right) - 1</math>, which is easier to compute. Either way, plugging in <math>n=29.5</math> gives <math>\boxed{869}</math>. | |
+ | |||
+ | == Solution 5 (Prime Factorization) == | ||
+ | We have <math>(31)(30)(29)(28)+1=755161.</math> Since the alternating sum of the digits <math>7-5+5-1+6-1=11</math> is divisible by <math>11,</math> we conclude that <math>755161</math> is divisible by <math>11.</math> | ||
+ | |||
+ | We evaluate the original expression by prime factorization: | ||
+ | <cmath>\begin{align*} | ||
+ | \sqrt{(31)(30)(29)(28)+1}&=\sqrt{755161} \\ | ||
+ | &=\sqrt{11\cdot68651} \\ | ||
+ | &=\sqrt{11^2\cdot6241} \\ | ||
+ | &=\sqrt{11^2\cdot79^2} \\ | ||
+ | &=11\cdot79 \\ | ||
+ | &=\boxed{869}. | ||
+ | \end{align*}</cmath> | ||
+ | ~Vrjmath (Fundamental Logic) | ||
+ | |||
+ | ~MRENTHUSIASM (Reconstruction) | ||
− | == Solution 6 ( | + | == Solution 6 (Observation) == |
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>. | 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>. | ||
Latest revision as of 00:12, 8 October 2021
Contents
Problem
Compute .
Solution 1 (Symmetry)
Note that the four numbers to multiply are symmetric with the center at .
Multiply the symmetric pairs to get
and
.
.
Solution 2 (Symmetry)
Notice that . Then we can notice that
and that
. Therefore,
. This is because we have that
as per the equation
.
~qwertysri987
Solution 3 (Symmetry with Generalization)
More generally, we can prove that one more than the product of four consecutive integers must be a perfect square:
At
we have
~Novus677 (Fundamental Logic)
~MRENTHUSIASM (Reconstruction)
Solution 4 (Symmetry with Generalization)
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
.
Solution 5 (Prime Factorization)
We have Since the alternating sum of the digits
is divisible by
we conclude that
is divisible by
We evaluate the original expression by prime factorization:
~Vrjmath (Fundamental Logic)
~MRENTHUSIASM (Reconstruction)
Solution 6 (Observation)
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
.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.