Difference between revisions of "1989 AIME Problems/Problem 1"

Latest revision as of 23:12, 7 October 2021

Problem

Compute $\sqrt{(31)(30)(29)(28)+1}$ .

Solution 1 (Symmetry)

Note that the four numbers to multiply are symmetric with the center at $29.5$ . Multiply the symmetric pairs to get $31\cdot 28=868$ and $30\cdot 29=870$ . $\sqrt{868\cdot 870 + 1} = \sqrt{(869-1)(869+1) + 1} = \sqrt{869^2 - 1^2 + 1} = \sqrt{869^2} = \boxed{869}$ .

Solution 2 (Symmetry)

Notice that $(a+1)^2 = a \cdot (a+2) +1$ . Then we can notice that $30 \cdot 29 =870$ and that $31 \cdot 28 = 868$ . Therefore, $\sqrt{(31)(30)(29)(28) +1} = \sqrt{(870)(868) +1} = \sqrt{(868 +1)^2} = \boxed{869}$ . This is because we have that $a=868$ as per the equation $(a+1)^2 = a \cdot (a+2) +1$ .

~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: $\begin{align*} (a+3)(a+2)(a+1)(a)+1 &= [(a+3)(a)][(a+2)(a+1)]+1 \\ &= [a^2+3a][a^2+3a+2]+1 \\ &= [a^2+3a]^2+2[a^2+3a]+1 \\ &= [a^2+3a+1]^2. \end{align*}$ At $a=28,$ we have $\sqrt{(a+3)(a+2)(a+1)(a)+1}=a^2+3a+1=\boxed{869}.$ ~Novus677 (Fundamental Logic)

~MRENTHUSIASM (Reconstruction)

Solution 4 (Symmetry with Generalization)

Similar to Solution 1 above, call the consecutive integers $\left(n-\frac{3}{2}\right), \left(n-\frac{1}{2}\right), \left(n+\frac{1}{2}\right), \left(n+\frac{3}{2}\right)$ to make use of symmetry. Note that $n$ itself is not an integer - in this case, $n = 29.5$ . The expression becomes $\sqrt{\left(n-\frac{3}{2}\right)\left(n + \frac{3}{2}\right)\left(n - \frac{1}{2}\right)\left(n + \frac{1}{2}\right) + 1}$ . Distributing each pair of difference of squares first, and then distributing the two resulting quadratics and adding the constant, gives $\sqrt{n^4 - \frac{5}{2}n^2 + \frac{25}{16}}$ . The inside is a perfect square trinomial, since $b^2 = 4ac$ . It's equal to $\sqrt{\left(n^2 - \frac{5}{4}\right)^2}$ , which simplifies to $n^2 - \frac{5}{4}$ . You can plug in the value of $n$ from there, or further simplify to $\left(n - \frac{1}{2}\right)\left(n + \frac{1}{2}\right) - 1$ , which is easier to compute. Either way, plugging in $n=29.5$ gives $\boxed{869}$ .

Solution 5 (Prime Factorization)

We have $(31)(30)(29)(28)+1=755161.$ Since the alternating sum of the digits $7-5+5-1+6-1=11$ is divisible by $11,$ we conclude that $755161$ is divisible by $11.$

We evaluate the original expression by prime factorization: $\begin{align*} \sqrt{(31)(30)(29)(28)+1}&=\sqrt{755161} \\ &=\sqrt{11\cdot68651} \\ &=\sqrt{11^2\cdot6241} \\ &=\sqrt{11^2\cdot79^2} \\ &=11\cdot79 \\ &=\boxed{869}. \end{align*}$ ~Vrjmath (Fundamental Logic)

~MRENTHUSIASM (Reconstruction)

Solution 6 (Observation)

The last digit under the radical is $1$ , so the square root must either end in $1$ or $9$ , since $x^2 = 1\pmod {10}$ means $x = \pm 1$ . Additionally, the number must be near $29 \cdot 30 = 870$ , narrowing the reasonable choices to $869$ and $871$ .

Continuing the logic, the next-to-last digit under the radical is the same as the last digit of $28 \cdot 29 \cdot 3 \cdot 31$ , which is $6$ . Quick computation shows that $869^2$ ends in $61$ , while $871^2$ ends in $41$ . Thus, the answer is $\boxed{869}$ .

@@ Line 2: / Line 2: @@
 Compute <math>\sqrt{(31)(30)(29)(28)+1}</math>.
-== Solution ==
+== Solution 1 (Symmetry)==
-=== Solution 1===
-Notice <math>{31*28 = 868}</math> and <math>{30*29 =870}</math>. So now our expression is <math>\sqrt{(870)(868) + 1}</math>. Setting 870 equal to <math>y</math>, we get <math>\sqrt{(y-1)^{2}}</math> which then equals <math>{(y-1)}</math>. So since <math>{y = 870}</math>, <math>{y-1}=869</math>, our answer is <math>\boxed{869}</math>.
-=== 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>.
 <math>\sqrt{868\cdot 870 + 1} = \sqrt{(869-1)(869+1) + 1} = \sqrt{869^2 - 1^2 + 1} = \sqrt{869^2} = \boxed{869}</math>.
-=== Solution 3===
+== Solution 2 (Symmetry)==
-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>.
+Notice that <math>(a+1)^2 = a \cdot (a+2) +1</math>. Then we can notice that <math>30 \cdot 29 =870 </math> and that <math>31 \cdot 28 = 868</math>. Therefore, <math> \sqrt{(31)(30)(29)(28) +1} = \sqrt{(870)(868) +1} = \sqrt{(868 +1)^2} = \boxed{869}</math>. This is because we have that <math>a=868</math> as per the equation <math>(a+1)^2 = a \cdot (a+2) +1</math>.
+~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:
+<cmath>\begin{align*}
+(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>
+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)
-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>.
+~MRENTHUSIASM (Reconstruction)
-===Solution 4===
+== 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===
+== Solution 5 (Prime Factorization) ==
-Multiplying <math>(31)(30)(29)(28)</math> gives us <math>755160</math>. Adding <math>1</math> to this gives <math>755161</math>. Now we must choose a number squared that is equal to <math>755161</math>. Taking the square root of this gives <math>\boxed{869}</math>
+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 (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>.
+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>.
 == 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

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