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

(→‎Solution 6)
(I title all solutions, remove duplicated solutions, and rearrange the solutions based on education values. If anyone disagrees, I am happy to discuss these edits in private-messages. Hope that I won't offend anyone.)
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Compute <math>\sqrt{(31)(30)(29)(28)+1}</math>.
 
Compute <math>\sqrt{(31)(30)(29)(28)+1}</math>.
  
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== Solution ==
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== Solution 1 (Symmetry)==
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=== Solution 1===
 
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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>.
 
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=== Solution 2===
 
 
Note that the four numbers to multiply are symmetric with the center at <math>29.5</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>.
 
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>.
 
<math>\sqrt{868\cdot 870 + 1} = \sqrt{(869-1)(869+1) + 1} = \sqrt{869^2 - 1^2 + 1} = \sqrt{869^2} = \boxed{869}</math>.
  
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=== Solution 3===
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== Solution 2 (Symmetry)==
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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>.
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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>.
  
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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>.
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~qwertysri987
  
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===Solution 4===
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== Solution 3 (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>.
 
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>.
  
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===Solution 5===
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== Solution 4 (Symmetry with Generalization) ==
 
Note that <math>a(a+1)(a+2)(a+3)+1=(a^2+3a+1)^2</math>.
 
Note that <math>a(a+1)(a+2)(a+3)+1=(a^2+3a+1)^2</math>.
 
So, our answer is just <math>28^2+3\cdot 28+1=\boxed{869}</math>
 
So, our answer is just <math>28^2+3\cdot 28+1=\boxed{869}</math>
  
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===Solution 6===
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== Solution 5 (Prime Factorizations) ==
 
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>
 
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>
  
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===Solution 7===
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== Solution 6 (Observations) ==
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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>.
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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>.
  
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~qwertysri987
+
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 ==
 
== See also ==

Revision as of 12:24, 28 June 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)

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 4 (Symmetry with Generalization)

Note that $a(a+1)(a+2)(a+3)+1=(a^2+3a+1)^2$. So, our answer is just $28^2+3\cdot 28+1=\boxed{869}$

Solution 5 (Prime Factorizations)

Multiplying $(31)(30)(29)(28)$ gives us $755160$. Adding $1$ to this gives $755161$. Now we must choose a number squared that is equal to $755161$. Taking the square root of this gives $\boxed{869}$

Solution 6 (Observations)

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 $\boxed{869}$.

See also

1989 AIME (Problems • Answer Key • Resources)
Preceded by
First Question 		Followed by
Problem 2
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All AIME Problems and Solutions

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