Difference between revisions of "2006 AMC 10B Problems/Problem 25"
(→Solution 2) |
(→Solution 2) |
||
Line 12: | Line 12: | ||
== Solution 2 == | == Solution 2 == | ||
− | 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 or X0X0</math> (X being the other digit in the license plate) . The condition <math>XX00</math> is impossible as Mr. Jones can't be 00 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 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 or X0X0</math> (X being the other digit in the license plate) . The condition <math>XX00</math> is impossible as Mr. Jones can't be 00 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 18:40, 1 January 2014
Contents
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?
Solution
Let be the set of the ages of Mr. Jones' children (in other words
if Mr. Jones has a child who is
years old). Then
and
. Let
be the positive integer seen on the license plate. Since at least one of
or
is contained in
, we have
.
We would like to prove that , so for the sake of contradiction, let us suppose that
. Then
so the units digit of
is
. Since the number has two distinct digits, each appearing twice, another digit of
must be
. Since Mr. Jones can't be
years old, the last two digits can't be
. Therefore
must be of the form
, where
is a digit. Since
is divisible by
, the sum of the digits of
must be divisible by
(see Divisibility rules for 9). Hence
which implies
. But
is not divisible by
, contradiction. So
and
is not the age of one of Mr. Jones' kids.
(We might like to check that there does, indeed, exist such a positive integer . If
is not an age of one of Mr. Jones' kids, then the license plate number must be a multiple of
. Since
and
is the only
digit multiple of
that fits all the conditions of the license plate's number, the license plate's number is
.)
Solution 2
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 . This means that the license plate would have to have two
digits, and would either be of the form
(X being the other digit in the license plate) . The condition
is impossible as Mr. Jones can't be 00 years old. If we separate the second condition,
into its prime factors, we get
.
and
are prime, therefore can't account for allowing
being evenly divisible by the childrens' ages. This leaves
, but no single digit number contains all the prime factors of the childrens ages, thus
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.