| The Problem: |
 |
Can the sixteen digits
2,2,3,3,4,4,5,5,6,6,7,7,8,8,9,9
be arranged to form two
8-digit numbers A and B with B=2A ?
Remarks. You might wish to practice by arranging the digits 1,1,2,2,4,4,5,5,7,7,8,8 into two 6-digit numbers C and D with D=2C , but please do not submit the
solution for this practice problem; and please do not submit any long solution
to our monthly problem that involves intricate case
analyses. |
Solution: No.
Solution 1.
(For this solution one
must know that any number is congruent modulo 3 to the sum of its digits.) For
any arrangement of the given digits that produces two numbers A and B , we see that
A+B≡≡≡ (sum of the digits of A+B)
(sum of the digits of A) +
(sum of the digits of B)2⋅(2+3+4+5+6+7+8+9)≡2⋅2≡1(mod 3).
If, however,
A and
B were to satisfy
B=2A , then
A+B=A+2A=3A≡0(mod 3).
Since both congruences cannot be satisfied simultaneously, the answer is
no, the given sixteen digits cannot be arranged to form two numbers
A and
B with
B=2A .
Solution
2. We shall present Lamis Alsheikh's solution, which in some ways was
the simplest of the submissions that avoided modular arithmetic. To follow his
argument you might want first to look at the suggested practice problem: to
arrange the digits
1,1,2,2,4,4,5,5,7,7,8,8 into two 6-digit numbers
C and
D with
D=2C . You perhaps recognize that these
are the digits in one period of the decimal expansion of both 1/7 and 2/7, so
C=142857 serves as a solution (with
D=2C=285714 ). Note that the 5 in
D comes from having multiplied the 2
in
C by 2 — there is a carry of 1 from
the digit greater than 4 to the right of 2; the 4 in
D comes from the 7 in
C because there is no carry involved.
What is important here is that the even digits in
D come from the digits of
C with a number less than 5 on the
right, while the odd digits in
D come from the digits of
C with a number 5 or greater on the
right. It turns out that there are 12 arrangements of the digits
1,2,4,5,7,8 into candidates for
C for which
2C is a permutation of those digits.
In each such arrangement, 2, 8, and 5 have a digit 5 or greater on the right, so
their double produces an odd digit of
2C . For another example, note that
the digits 1 through 8 can be arranged into
8 -digit numbers whose digits are
permuted by multiplication by 2;
E=42715638 is such an example. We return now
to our March problem.
Suppose that the 16 given digits can be arranged
into numbers
A and
B for which
B=2A . Denote by
an and
bn the number of
times the digit
n appears in the numbers
A and
B , respectively. Then for each of
the eight different values of
n we have
Because 2 times the digit 5 in
A would produce either a 1 or 0 in
B (according as there is or is not a
carry), both of the 5's must appear in
B :
Similarly, because any 2 or 3 in
B would necessarily come from a 6 in
A , we have
a6=b2+b3 . Adding
b5+b6 to both sides
of this equation yields (with the help of (1) and (2))
b5+(b6+a6)=4=b2+b3+b5+b6,
whence (because
B consists of eight digits)
Because any digits 8 or 9 in
B would come from 4 or 9 in
A , we have
a4+a9=b8+b9 , whence
(adding
b4 to both sides
and invoking (3))
2+a9==b4+b8+b94−b7=4−(2−a7)=2+a7.
Thus,
observing the correspondence between digits greater than 4 in
A and odd digits in
B , we have
From (1), (2), and (4) this equation becomes
whence
We deduce immediately from (5) that
b7≠2 . Nor can
b7 equal 0:
otherwise (2) and (5) would imply that
b3=b6=b8=b5=2 , which in turn implies that both
a2 and
a7 would have to
be 2, so that
b4 and
b5 would have to
be 2, which would force five of the
b 's to be 2, a contradiction. The
remaining possibility is that
a7=b7=a9=b9=1 . But this also leads to a
contradiction: Equation (5) then becomes
Because any digits 6 and 7 in
B come from 3 and 8 in
A ,
b6+b7=a3+a8 . Setting
b7=1 we get
b6+1=4−(b3+b8)=4−(2−b6)=2+b6,
which is a contradiction. We conclude that there can be no such
A and
B .
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