Difference between revisions of "2006 AMC 10B Problems/Problem 25"
(→Solution 4) |
(→Solution 4) |
||
Line 22: | Line 22: | ||
Since 5 has the "strictest" divisibility rules of the single digit numbers, we check that case first. To be divisible by 9, the other repeating digit must be 4. Seeing that 5544 is the only arrangement of 2 4's and 2 5's that is divisible by the other single digit numbers and itself is not divisible by 5, the answer is <math>\boxed {(B) 5}</math>. | Since 5 has the "strictest" divisibility rules of the single digit numbers, we check that case first. To be divisible by 9, the other repeating digit must be 4. Seeing that 5544 is the only arrangement of 2 4's and 2 5's that is divisible by the other single digit numbers and itself is not divisible by 5, the answer is <math>\boxed {(B) 5}</math>. | ||
-liu4505 | -liu4505 | ||
+ | -edited by mobius247 | ||
== See also == | == See also == |
Revision as of 21:45, 2 June 2021
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?
Solutions
Solution 1
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's 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 Mr. Jones' children is of the age 5, then the license plate will have to end in the digit . The license plate cannot end in the digit
as
is a factor of the number, so it must be even. This means that the license plate would have to have two
digits, and would either be of the form
or
(X being the other digit in the license plate) . The condition
is impossible as Mr. Jones can't be
years old. If we separate the second condition,
into its prime factors, we get
.
is prime, and therefore can't account for allowing
being evenly divisible by the childrens' ages. The
accounts for the 5 and one factor of 2. This leaves
, but no single digit number contains all the prime factors besides 5 and 2 of the childrens' ages, thus
can't be one of the childrens' ages.
Solution 3
Another way to do the problem is by the process of elimination. The only possible correct choices are the highest powers of each prime, ,
,
, and
, since indivisibility by any powers lower than these means indivisibility by a higher power of the prime (for example, indivisibility by
means indivisibility by
. Since the number is divisible by
, it is not the answer, and the digits have to add up to
. Let
be the digits: since
,
. The only possible choices for
are
,
,
,
, and
. Since the last 2 digits are divisible by
,
,
, and
can be eliminated. Only
and
remain; since there is no
or
in either pair, the number cannot be divisible by
.
Solution 4
Since 5 has the "strictest" divisibility rules of the single digit numbers, we check that case first. To be divisible by 9, the other repeating digit must be 4. Seeing that 5544 is the only arrangement of 2 4's and 2 5's that is divisible by the other single digit numbers and itself is not divisible by 5, the answer is .
-liu4505
-edited by mobius247
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.