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Difference between revisions of "2006 AMC 12B Problems/Problem 19"

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== Problem ==
 
== Problem ==
 
Mr. Jones has eight children of different ages. On a family trip his oldest child, who is 9, spots a license plate with a 4-digit number in which each of two digits appears two times. "Look, daddy!" she exclaims. "That number is evenly divisible by the age of each of us kids!" "That's right," replies Mr. Jones, "and the last two digits just happen to be my age." Which of the following is not the age of one of Mr. Jones's children?
 
Mr. Jones has eight children of different ages. On a family trip his oldest child, who is 9, spots a license plate with a 4-digit number in which each of two digits appears two times. "Look, daddy!" she exclaims. "That number is evenly divisible by the age of each of us kids!" "That's right," replies Mr. Jones, "and the last two digits just happen to be my age." Which of the following is not the age of one of Mr. Jones's children?
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== Solution ==
 
== Solution ==
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===Solution 1===
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First, The number of the plate is divisible by <math>9</math> and in the form of
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First, The number of the plate is divisible by 9 and in the form of
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<math>aabb</math>, <math>abba</math> or <math>abab</math>.
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aabb, abba or abab
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We can conclude straight away that a+b is 9 using the 9 divisibility rule
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We can conclude straight away that <math>a+b= 9</math> using the <math>9</math> divisibility rule.
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After trying various numbers, and reaching five
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I can observe that b=0 or 5.
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If <math>b=1</math>, the number is not divisible by <math>2</math> (unless it's <math>1818</math>, which is not divisible by <math>4</math>), which means there are no <math>2</math>, <math>4</math>, <math>6</math>, or <math>8</math> year olds on the car, but that can't be true, as that would mean there are less than <math>8</math> kids on the car.
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If b=5, the number is not divisible by 2
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If b=0, the number must be 9900, and is not divisible by 8
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If <math>b=2</math>, then the only possible number is <math>7272</math>. <math>7272</math> is divisible by <math>4</math>, <math>6</math>, and <math>8</math>, but not by <math>5</math> and <math>7</math>, so that doesn't work.  
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Answer B 5
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If <math>b=3</math>, then the only number is <math>6336</math>, also not divisible by <math>5</math> or <math>7</math>.
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If <math>b=4</math>, the only number is <math>5544</math>. It is divisible by <math>4</math>, <math>6</math>, <math>7</math>, and <math>8</math>.
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Therefore, we conclude that the answer is <math> \mathrm{(B)}\ 5 </math>
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'''NOTE''': Automatically, since there are 8 children and all of their ages are less than or equal to 9 and are different, the answer choices can be narrowed down to <math>5</math> or <math>8</math>.
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== Alternate Solution ==
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We know that the number of the plate is divisible by <math>9</math> and in the form <math>aabb</math>, <math>abba</math>, or <math>abab</math> for distinct digits <math>a,b</math>.
 +
 
 +
Using the divisibility rule for <math>9</math>, we can conclude that <math>a+b\equiv 0\pmod{9}</math>.
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We also know that the number of the plate is even, because you can only discard one number from the integers <math>1</math> through <math>9</math>, inclusive (<math>8</math> children, oldest is <math>9</math>), and there's always going to be an even number left.
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If one of the children was <math>5</math> years old, then the plate number would be divisible by both <math>5</math> and <math>2</math>. Thus, the units digit must be <math>0</math>.
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Then, the possible form of the plate number would be <math>aa00</math>, <math>0bb0</math>, or <math>a0a0</math>. The first case is not possible because <math>00</math> is not a possible age for the father.
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We have the remaining forms <math>0bb0</math> and <math>a0a0</math>.
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<math>\textbf{Case 1: 0bb0:}</math>
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We know that <math>2b\equiv 0\pmod{9}</math>, so <math>b</math> has to be either <math>0</math> or <math>9</math>. <math>b</math> can't be <math>0</math> because <math>a, b</math> are distinct. So, <math>b=9</math>. However, the number <math>0990</math> is not divisible by both <math>4</math> and <math>7</math>, so whichever number you discard, the other one will still be there.  This creates a contradiction, so Case <math>1</math> cannot be true.
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<math>\textbf{Case 2: a0a0:}</math>
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Similarly to Case <math>1</math>, we can determine that <math>a</math> has to be <math>9</math>. However, the number <math>9090</math> is not divisible by both <math>4</math> and <math>7</math>. Using the same logic in case <math>1</math>, we conclude that Case <math>2</math> cannot be true.
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Since we have disproven all of our cases, we know that it is impossible for one of the children to be <math> \mathrm{(B)}\ 5 </math> years old.
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'''NOTE''': This might look tedious, but it only took me around 30 seconds to do on paper.
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2006|ab=B|num-b=18|num-a=20}}
 
{{AMC12 box|year=2006|ab=B|num-b=18|num-a=20}}
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{{MAA Notice}}

Latest revision as of 08:47, 10 August 2015

Problem

Mr. Jones has eight children of different ages. On a family trip his oldest child, who is 9, spots a license plate with a 4-digit number in which each of two digits appears two times. "Look, daddy!" she exclaims. "That number is evenly divisible by the age of each of us kids!" "That's right," replies Mr. Jones, "and the last two digits just happen to be my age." Which of the following is not the age of one of Mr. Jones's children?

$\mathrm{(A)}\ 4\qquad\mathrm{(B)}\ 5\qquad\mathrm{(C)}\ 6\qquad\mathrm{(D)}\ 7\qquad\mathrm{(E)}\ 8$

Solution

First, The number of the plate is divisible by $9$ and in the form of $aabb$, $abba$ or $abab$.

We can conclude straight away that $a+b= 9$ using the $9$ divisibility rule.

If $b=1$, the number is not divisible by $2$ (unless it's $1818$, which is not divisible by $4$), which means there are no $2$, $4$, $6$, or $8$ year olds on the car, but that can't be true, as that would mean there are less than $8$ kids on the car.

If $b=2$, then the only possible number is $7272$. $7272$ is divisible by $4$, $6$, and $8$, but not by $5$ and $7$, so that doesn't work.

If $b=3$, then the only number is $6336$, also not divisible by $5$ or $7$.

If $b=4$, the only number is $5544$. It is divisible by $4$, $6$, $7$, and $8$.

Therefore, we conclude that the answer is $\mathrm{(B)}\ 5$

NOTE: Automatically, since there are 8 children and all of their ages are less than or equal to 9 and are different, the answer choices can be narrowed down to $5$ or $8$.

Alternate Solution

We know that the number of the plate is divisible by $9$ and in the form $aabb$, $abba$, or $abab$ for distinct digits $a,b$.

Using the divisibility rule for $9$, we can conclude that $a+b\equiv 0\pmod{9}$.

We also know that the number of the plate is even, because you can only discard one number from the integers $1$ through $9$, inclusive ($8$ children, oldest is $9$), and there's always going to be an even number left.

If one of the children was $5$ years old, then the plate number would be divisible by both $5$ and $2$. Thus, the units digit must be $0$. Then, the possible form of the plate number would be $aa00$, $0bb0$, or $a0a0$. The first case is not possible because $00$ is not a possible age for the father.

We have the remaining forms $0bb0$ and $a0a0$.

$\textbf{Case 1: 0bb0:}$ We know that $2b\equiv 0\pmod{9}$, so $b$ has to be either $0$ or $9$. $b$ can't be $0$ because $a, b$ are distinct. So, $b=9$. However, the number $0990$ is not divisible by both $4$ and $7$, so whichever number you discard, the other one will still be there. This creates a contradiction, so Case $1$ cannot be true.

$\textbf{Case 2: a0a0:}$ Similarly to Case $1$, we can determine that $a$ has to be $9$. However, the number $9090$ is not divisible by both $4$ and $7$. Using the same logic in case $1$, we conclude that Case $2$ cannot be true.

Since we have disproven all of our cases, we know that it is impossible for one of the children to be $\mathrm{(B)}\ 5$ years old.

NOTE: This might look tedious, but it only took me around 30 seconds to do on paper.

See also

2006 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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