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Difference between revisions of "2006 AMC 10B Problems/Problem 25"

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== Solution 2 ==
 
== Solution 2 ==
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Alternatively, we can see that if one of Jones' children is of the age 5, then the license plate will have to end in the digit <math>0</math> . This means that the license plate would have to have two <math>0</math> digits, and would either be of the form <math>XX00</math> or <math>X0X0</math> (X being the other digit in the license plate) . The condition <math>XX00</math> is impossible as Mr. Jones can't be <math>00</math> years old. If we separate the second condition, <math>X0X0</math> into its prime factors, we get <math> X\cdot 11\cdot 101 </math>. <math>11</math> and <math>101</math> are prime, therefore can't account for allowing <math>X0X0</math> being evenly divisible by the childrens' ages. This leaves <math>X</math>, but no single digit number contains all the prime factors of the childrens ages, thus <math>5</math> can't be one of the childrens ages.
+
Alternatively, we can see that if one of Mr.Jones' children is of the age 5, then the license plate will have to end in the digit <math>0</math> . This means that the license plate would have to have two <math>0</math> digits, and would either be of the form <math>XX00</math> or <math>X0X0</math> (X being the other digit in the license plate) . The condition <math>XX00</math> is impossible as Mr. Jones can't be <math>00</math> years old. If we separate the second condition, <math>X0X0</math> into its prime factors, we get <math> X\cdot 11\cdot 101 </math>. <math>11</math> and <math>101</math> are prime, therefore can't account for allowing <math>X0X0</math> being evenly divisible by the childrens' ages. This leaves <math>X</math>, but no single digit number contains all the prime factors of the childrens ages, thus <math>5</math> can't be one of the childrens ages.
  
 
== See also ==
 
== See also ==

Revision as of 17:41, 1 January 2014

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

Let $S$ be the set of the ages of Mr. Jones' children (in other words $i \in S$ if Mr. Jones has a child who is $i$ years old). Then $|S| = 8$ and $9 \in S$. Let $m$ be the positive integer seen on the license plate. Since at least one of $4$ or $8$ is contained in $S$, we have $4 | m$.

We would like to prove that $5 \not\in S$, so for the sake of contradiction, let us suppose that $5 \in S$. Then $5\cdot 4 = 20 | m$ so the units digit of $m$ is $0$. Since the number has two distinct digits, each appearing twice, another digit of $m$ must be $0$. Since Mr. Jones can't be $00$ years old, the last two digits can't be $00$. Therefore $m$ must be of the form $d0d0$, where $d$ is a digit. Since $m$ is divisible by $9$, the sum of the digits of $m$ must be divisible by $9$ (see Divisibility rules for 9). Hence $9 | 2d$ which implies $d = 9$. But $m = 9090$ is not divisible by $4$, contradiction. So $5 \not\in S$ and $5$ is not the age of one of Mr. Jones' kids. $\boxed{ B }$

(We might like to check that there does, indeed, exist such a positive integer $m$. If $5$ is not an age of one of Mr. Jones' kids, then the license plate number must be a multiple of $lcm(1,2,3,4,6,7,8,9) = 504$. Since $11\cdot504 = 5544$ and $5544$ is the only $4$ digit multiple of $504$ that fits all the conditions of the license plate's number, the license plate's number is $5544$.)

Solution 2

Alternatively, we can see that if one of Mr.Jones' children is of the age 5, then the license plate will have to end in the digit $0$ . This means that the license plate would have to have two $0$ digits, and would either be of the form $XX00$ or $X0X0$ (X being the other digit in the license plate) . The condition $XX00$ is impossible as Mr. Jones can't be $00$ years old. If we separate the second condition, $X0X0$ into its prime factors, we get $X\cdot 11\cdot 101$. $11$ and $101$ are prime, therefore can't account for allowing $X0X0$ being evenly divisible by the childrens' ages. This leaves $X$, but no single digit number contains all the prime factors of the childrens ages, thus $5$ can't be one of the childrens ages.

See also

2006 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Last question
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 10 Problems and Solutions
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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