|
I Fundamentals, 4. Digits of Numbers
From (2) and (3) it follows that
9
|
a
+
b
−
(
S
(
a
)
+
S
(
b
))
;
(
5
)
hence
9
|
S
(
a
)
+
S
(
b
)
−
S
(
a
+
b
)
=
n
+
n
−
n
=
n
,
(
6
)
as desired.
(b) Conversely, we prove that if
n
=
9
p
is a multiple of 9, then integers
a
,
b
>
0 with
S
(
a
)
=
S
(
b
)
=
S
(
a
+
b
)
can be found. Indeed, set
a
=
531531
. . .
531
3
p
digits
and
b
=
171171
. . .
171
3
p
digits
. Then
a
+
b
=
702702
. . .
702
3
p
digits
and
S
(
a
)
=
S
(
b
)
=
S
(
a
+
b
)
=
9
p
=
n
,
as claimed.
Additional Problems
Problem 4.2.7.
Show that there exist infinitely many natural numbers
n
such that
S
(
3
n
)
≥
S
(
3
n
+
1
)
.
(1997 Russian Mathematical Olympiad)
Problem 4.2.8.
Do there exist three natural numbers
a
,
b
,
c
such that
S
(
a
+
b
) <
5,
S
(
b
+
c
) <
5,
S
(
c
+
a
) <
5, but
S
(
a
+
b
+
c
) >
50?
(1998 Russian Mathematical Olympiad)
Problem 4.2.9.
Prove that there exist distinct positive integers
{
n
i
}
1
≤
i
≤
50
such
that
n
1
+
S
(
n
1
)
=
n
2
+
S
(
n
2
)
= · · · =
n
50
+
S
(
n
50
).
(1999 Polish Mathematical Olympiad)
Problem 4.2.10.
The sum of the decimal digits of the natural number
n
is 100,
and that of 44
n
is 800. What is the sum of the digits of 3
n
?
(1999 Russian Mathematical Olympiad)
Problem 4.2.11.
Consider all numbers of the form 3
n
2
+
n
+
1, where
n
is a
positive integer.
(a) How small can the sum of the digits (in base 10) of such a number be?
(b) Can such a number have the sum of its digits (in base 10) equal to 1999?
(1999 United Kingdom Mathematical Olympiad)
4.3. Other Problems Involving Digits
85
Problem 4.2.12.
Consider the set
A
of all positive integers
n
with the following
properties: the decimal expansion contains no 0, and the sum of the (decimal)
digits of
n
divides
n
.
(a) Prove that there exist infinitely many elements in
A
with the following
property: the digits that appear in the decimal expansion of
A
appear the same
number of times.
(b) Show that for each positive integer
k
, there exists an element in
A
with
exactly
k
digits.
(2001 Austrian–Polish Mathematics Competition)
4.3
Other Problems Involving Digits
Problem 4.3.1.
Prove that there are at least
666
positive composite numbers with
2006
digits, having a digit equal to
7
and all the rest equal to
1
.
Solution.
The given numbers are
n
k
=
111
. . .
17 11
. . .
1
k
digits
=
111
. . .
1
2006 digits
+
6 000
. . .
0
k
digits
=
1
9
(
10
2006
−
1
)
+
6
·
10
k
,
k
=
0
,
1
, . . . ,
2005
.
It is obvious that none of these numbers is a multiple of 2, 3, 5, or 11, since
11 divides 111
. . .
1
2006 digits
, but not 6
·
10
k
.
So we are led to the idea of counting multiples of 7 and 13. We have 9
n
k
=
100
·
1000
668
−
1
+
54
·
10
k
≡
2
·
(
−
1
)
668
−
1
+
(
−
2
)
·
10
k
≡
1
−
2
·
10
k
(
mod 7
)
;
hence 7
|
n
k
if 10
k
≡
3
k
≡
4
(
mod 7
)
. This happens for
k
=
4
,
10
,
16
, . . . ,
2002,
so there are 334 multiples of 7. Furthermore, 9
n
k
≡
7
·
(
−
1
)
668
−
1
+
2
·
10
k
=
6
+
2
·
10
k
(
mod 13
)
; hence 13
|
n
k
if 10
k
≡
10
(
mod 13
)
. This happens for
k
=
1
,
7
,
13
,
19
, . . . ,
2005, so there are 335 multiples of 13. In all we have found
669 nonprime numbers.
Problem 4.3.2.
Let a
1
,
a
2
, . . . ,
a
10
6
be integers between
1
and
9
, inclusive. Prove
that at most
100
of the numbers a
1
a
2
· · ·
a
k
(
1
≤
k
≤
10
6
)
are perfect squares.
(2001 Russian Mathematical Olympiad)
Solution.
For each positive integer
x
, let
d
(
x
)
be the number of decimal digits
in
x
.
Lemma.
Suppose that y
>
x are perfect squares such that y
=
10
2
b
x
+
c for
some positive integers b
,
c with c
<
10
2
b
. Then
d
(
y
)
−
1
≥
2
(
d
(
x
)
−
1
).
86
I Fundamentals, 4. Digits of Numbers
Proof.
Because
y
>
10
2
b
x
, we have
√
y
>
10
b
√
x
. Because
√
y
and 10
b
√
x
are
both integers,
√
y
≥
10
b
√
x
+
1, so that 10
2
b
x
+
c
=
y
≥
10
2
b
x
+
2
·
10
b
√
x
+
1.
Thus,
c
≥
2
·
10
b
√
x
+
1.
Also, 10
2
b
>
c
by assumption, implying that
10
2
b
>
c
≥
2
·
10
b
√
x
+
1
.
Hence, 10
b
>
2
√
x
. It follows that
y
>
10
2
b
x
>
4
x
2
.
Therefore,
d
(
y
)
≥
2
d
(
x
)
−
1
,
as desired.
We claim that there are at most 20 perfect squares
a
1
a
2
· · ·
a
k
with an even
(resp. odd) number of digits. Let
s
1
<
s
2
<
· · ·
<
s
n
be these perfect squares.
Clearly
d
(
s
n
)
≤
10
6
. We now prove that if
n
>
1, then
d
(
s
n
)
≥
1
+
2
n
−
1
.
Because
s
1
,
s
2
, . . . ,
s
n
all have an even (resp. odd) number of digits, for each
i
=
1
,
2
, . . . ,
n
−
1, we can write
s
i
+
1
=
10
2
b
s
i
+
c
for some integers
b
>
0 and
0
≤
c
<
10
2
b
. Because no
a
i
equals 0, we further know that 0
<
c
. Hence, by our
lemma,
d
(
s
i
+
1
)
−
1
≥
2
(
d
(
s
i
)
−
1
)
for each
i
=
1
,
2
, . . . ,
n
−
1. Because
d
(
s
2
)
−
1
≥
2, we thus have
d
(
s
n
)
−
1
≥
2
n
−
1
, as desired.
Thus, if
n
>
1,
1
+
2
n
−
1
≤
d
(
s
n
)
≤
10
6
,
and
n
≤
#
log
(
10
6
−
1
)
log 2
$
+
1
=
20
.
Hence, there are at most 20 perfect squares
a
1
a
2
· · ·
a
k
with an even (resp.
odd) number of digits.
Therefore, there are at most 40
<
100 perfect squares
a
1
a
2
· · ·
a
k
.
Additional Problems
Problem 4.3.3.
A
wobbly
number is a positive integer whose digits in base 10 are
alternately nonzero and zero, the units digit being nonzero. Determine all positive
integers that do not divide any wobbly number.
(35th International Mathematical Olympiad Shortlist)
4.3. Other Problems Involving Digits
87
Problem 4.3.4.
A positive integer is called
monotonic
if its digits in base 10, read
from left right, are in nondecreasing order. Prove that for each
n
∈
N
, there exists
an
n
-digit monotonic number that is a perfect square.
(2000 Belarusian Mathematical Olympiad)
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