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Difference between revisions of "2013 AIME I Problems/Problem 2"

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== Solution==
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== Solution 1==
 
The number takes a form of <math>\overline{5xyz5}</math>, in which <math>5|(x+y+z)</math>. Let <math>x</math> and <math>y</math> be arbitrary digits. For each pair of <math>x,y</math>, there are exactly two values of <math>z</math> that satisfy the condition of <math>5|(x+y+z)</math>. Therefore, the answer is <math>10\times10\times2=\boxed{200}</math>
 
The number takes a form of <math>\overline{5xyz5}</math>, in which <math>5|(x+y+z)</math>. Let <math>x</math> and <math>y</math> be arbitrary digits. For each pair of <math>x,y</math>, there are exactly two values of <math>z</math> that satisfy the condition of <math>5|(x+y+z)</math>. Therefore, the answer is <math>10\times10\times2=\boxed{200}</math>
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== Solution 2 (casework)==
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We know the number will take the form <math>\overline{5xyz5}</math> because of the first two conditions. The third condition means that <math>5|(x+y+z)</math>, where <math>x,y,z</math> are nonnegative integers less than <math>10</math>. Let's split the problem into cases, where each case represents a possible sum of <math>x,y,z</math>.
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1. If <math>x+y+z=0</math>, we only have <math>x=0,y=0,z=0</math>, so only <math>1</math>.
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2. If <math>x+y+z=5</math>, we use stars & bars. We have <math>5</math> stars and <math>3-1=2</math> bars, so this case has <math>7\choose{2}</math><math> = 21</math>.
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3. If <math>x+y+z=10</math>, we use similar logic. We have <math>10</math> stars and <math>2</math> bars, so <math>12\choose{2}</math><math> = 66</math>. However, <math>x,y,z</math> must be less than <math>10</math>. Three of our order pairs have <math>10: (10,0,0), (0,10,0), (0,0,10)</math>. Therefore, this case has <math>66-3=63</math>.
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4. If <math>x+y+z=15</math>, it gets more complicated. Using the same system as used previously would be too complicated. But remember that this case is equivalentto if we wanted to choose <math>x+y+z=12</math>. We know this because you can take any ordered pair satisfying <math>x+y+z=15,</math> subtract each variable from <math>9</math>, and get an ordered pair satisfying <math>x+y+z=12</math>. For example, take <math>(5,6,4)</math>, which satisfies <math>x+y+z=15.</math> Its corresponding ordered pair would be <math>(4,3,5)</math>, which satisfies <math>x+y+z=12</math>. Let's proceed to calculating. Applying stars and bars, we get <math>14\choose{2}</math><math> = 91</math>, but we have to subtract the subcases including <math>10,11, </math> or <math>12</math> because <math>x,y,z</math> must all be one-digit integers. There are <math>3</math> cases with a <math>12</math>, <math>6</math> cases with an <math>11</math>, and <math>3+6</math> cases with a <math>10</math>. So this case has <math>91-18=73</math>.
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5. If <math>x+y+z=20</math>, we can just calculate the ways to get <math>x+y+z=7</math>. Applying stars & bars, we get <math>9\choose{2}</math><math> = 36</math>.
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6. If <math>x+y+z=25</math>, we can just calculate the ways to get <math>x+y+z=2.</math> Applying stars & bars, we get <math>4\choose{2}</math><math> = 6</math>.
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Therefore, our answer is <math>1+21+63+73+36+6 = \boxed{200}</math>.
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Note: For case 4, the subcases that must be excluded are <math>(12,0,0), (11,1,0), (10,2,0), (10,1,1)</math> and each of their respective permutations. Those <math>4</math> ordered pairs have <math>3,6,6,3</math> permutations respectively, which is why <math>18</math> ordered pairs must be subtracted from <math>91</math>.
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~lprado
  
 
==Video Solution==
 
==Video Solution==

Latest revision as of 20:00, 30 November 2024

Problem

Find the number of five-digit positive integers, $n$, that satisfy the following conditions:

    (a) the number $n$ is divisible by $5,$
    (b) the first and last digits of $n$ are equal, and
    (c) the sum of the digits of $n$ is divisible by $5.$


Solution 1

The number takes a form of $\overline{5xyz5}$, in which $5|(x+y+z)$. Let $x$ and $y$ be arbitrary digits. For each pair of $x,y$, there are exactly two values of $z$ that satisfy the condition of $5|(x+y+z)$. Therefore, the answer is $10\times10\times2=\boxed{200}$

Solution 2 (casework)

We know the number will take the form $\overline{5xyz5}$ because of the first two conditions. The third condition means that $5|(x+y+z)$, where $x,y,z$ are nonnegative integers less than $10$. Let's split the problem into cases, where each case represents a possible sum of $x,y,z$.

1. If $x+y+z=0$, we only have $x=0,y=0,z=0$, so only $1$.

2. If $x+y+z=5$, we use stars & bars. We have $5$ stars and $3-1=2$ bars, so this case has $7\choose{2}$$= 21$.

3. If $x+y+z=10$, we use similar logic. We have $10$ stars and $2$ bars, so $12\choose{2}$$= 66$. However, $x,y,z$ must be less than $10$. Three of our order pairs have $10: (10,0,0), (0,10,0), (0,0,10)$. Therefore, this case has $66-3=63$.

4. If $x+y+z=15$, it gets more complicated. Using the same system as used previously would be too complicated. But remember that this case is equivalentto if we wanted to choose $x+y+z=12$. We know this because you can take any ordered pair satisfying $x+y+z=15,$ subtract each variable from $9$, and get an ordered pair satisfying $x+y+z=12$. For example, take $(5,6,4)$, which satisfies $x+y+z=15.$ Its corresponding ordered pair would be $(4,3,5)$, which satisfies $x+y+z=12$. Let's proceed to calculating. Applying stars and bars, we get $14\choose{2}$$= 91$, but we have to subtract the subcases including $10,11,$ or $12$ because $x,y,z$ must all be one-digit integers. There are $3$ cases with a $12$, $6$ cases with an $11$, and $3+6$ cases with a $10$. So this case has $91-18=73$.

5. If $x+y+z=20$, we can just calculate the ways to get $x+y+z=7$. Applying stars & bars, we get $9\choose{2}$$= 36$.

6. If $x+y+z=25$, we can just calculate the ways to get $x+y+z=2.$ Applying stars & bars, we get $4\choose{2}$$= 6$.

Therefore, our answer is $1+21+63+73+36+6 = \boxed{200}$.

Note: For case 4, the subcases that must be excluded are $(12,0,0), (11,1,0), (10,2,0), (10,1,1)$ and each of their respective permutations. Those $4$ ordered pairs have $3,6,6,3$ permutations respectively, which is why $18$ ordered pairs must be subtracted from $91$.

~lprado

Video Solution

https://www.youtube.com/watch?v=kz3ZX4PT-_0 ~Shreyas S

See also

2013 AIME I (Problems • Answer Key • Resources)
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