30 Chapter 3
Practice
208 Problem Find an equivalent fraction for the repeating decimal 0
.3172.
209 Problem
A two-digit number is divided by the sum of its digits. What is the largest possible remainder?
210 Problem
Shew that the integer 11
. . . 11
| {z
}
221 1′ s
is a composite number.
211 Problem Let a and b be the integers
a = 111 . . . 1
| {z
}
m 1′s
b = 1 000 . . . 0
| {z
}
m−1 0′ s
5 .
Shew that ab + 1 is a perfect square.
212 Problem What digits appear on the product
3 . . . 3
|{z}
666 3′s
· 6... 6
|{z}
666 6′ s
?
213 Problem
Shew that there exist no integers with the following property: if the initial digit is suppressed, the resulting integer is 1
35 of the original number.
214 Problem
Shew that the sum of all the integers of n digits, n ≥ 3, is
494 99 . . . 9
| {z
}
n−3 9′ s
55 00 . . . 0
| {z
}
n−2 0′s
.
215 Problem Shew that for any positive integer n,
11 . . . 1
| {z
}
2n 1′ s
− 22 . . . 2
| {z
}
n 2′ s
is a perfect square.
216 Problem A whole number is equal to the arithmetic mean of all the numbers ob-
tained from the given number with the aid of all possible permutation of its digits. Find all whole numbers with that property.
217 Problem The integer n is the smallest multiple of 15 such that every digit of n is
either 0 or 8. Compute n
15 .
218 Problem Shew that Champernowne’s number
.12345678910111213141516171819202122 . . . , which is the sequence of natural numbers written after the decimal point, is irrational.
219 Problem
Given that 1
49 = 0
.020408163265306122448979591836734693877551, find the last thousand digits of
1 + 50 + 50
2
+ ··· + 50
999
.
220 Problem Let t be a positive real number. Prove that there is a positive integer n
such that the decimal expansion of nt contains a 7.
221 Problem AIME 1989 Suppose that n is a positive integer and d is a single digit in
base-ten. Find n if n
810 = 0
.d25d25d25d25d25 . . .
222 Problem AIME 1988
Find the smallest positive integer whose cube ends in 888.
223 Problem AIME 1986 In the parlour game, the “magician” asks one of the partici-
pants to think of a three-digit number abc, where a , b, c represent the digits of the number
in the order indicated. The magician asks his victim to form the numbers acb
, bac, cab, cba, to add these numbers and to reveal their sum N. If told the value of N, the magician can
identify abc. Play the magician and determine abc if N = 319.
224 Problem AIME 1988 For any positive integer k, let f
1
k denote the square of the sums of the digits of k
. For n ≥ 2, let f
n
k = f
1
f
n−1
k . Find f
1988
11 .
225 Problem IMO 1969 Determine all three-digit numbers N that are divisible by 11
and such that N
11 equals the sum of the squares of the digits of N.
226 Problem IMO 1962 Find the smallest natural number having the last digit 6 and if
this 6 is erased and put in from of the other digits, the resulting number is four times as large as the original number.
3.3 Non-decimal Scales
The fact that most people have ten fingers has fixed our scale of notation to the decimal. Given any positive integer r 1, we
can, however, express any number x in base r. If n is a positive integer, and r
1 is an integer, then n has the base-r representation n = a
+ a
1
r + a
2
r
2
+ ··· + a
k
r
k
, 0 ≤ a
t
≤ r − 1, a
k
6= 0, r
k
≤ n r
k+1
. We use the convention that we shall refer to a decimal number without referring to its base, and to a base-r number by using
the subindex
r
.
Non-decimal Scales 31
227 Example
Express the decimal number 5213 in base-seven. Solution: Observe that 5213
7
5
. We thus want to find 0 ≤ a , . . . , a
4
≤ 6,a
4
6= 0 such that 5213 = a
4
7
4
+ a
3
7
3
+ a
2
7
2
+ a
1
7 + a .
Dividing by 7
4
, we obtain 2+ proper fraction = a
4
+ proper fraction. This means that a
4
= 2 . Thus 5213 = 2 · 7
4
+ a
3
7
3
+ a
2
7
2
+ a
1
7 + a or 411 = 5213 = a
3
7
3
+ a
2
7
2
+ a
1
7 + a . Dividing by 7
3
this last equality we obtain 1+ proper fraction = a
3
+ proper fraction, and so a
3
= 1 . Continuing in this way we deduce that 5213 = 21125
7
. The method of successive divisions used in the preceding problem can be conveniently displayed as
7 5212
5 7
744 2
7 106
1 7
15 1
7 2
2 The central column contains the successive quotients and the rightmost column contains the corresponding remainders. Reading
from the last remainder up, we recover 5213 = 21125
7
.
228 Example Write 562
7
in base-five. Solution: 562
7
= 5 · 7
2
+ 6 · 7 + 2 = in decimal scale, so the problem reduces to convert 289 to base-five. Doing successive
divisions, 5
289 4
5 57
2 5
11 1
5 2
2 Thus 562
7
= 289 = 2124
5
.
229 Example Express the fraction
13 16
in base-six. Solution: Write
13 16
= a
1
6 +
a
2
6
2
+ a
3
6
3
+ a
4
6
4
+ ···
Multiplying by 6, we obtain 4+ proper fraction = a
1
+ proper fraction, so a
1
= 4 . Hence
13 16
− 4
6 =
7 48
= a
2
6
2
+ a
3
6
3
+ a
4
6
4
+ ···
Multiply by 6
2
we obtain 5+ proper fraction = a
2
+ proper fraction, and so a
2
= 5 . Continuing in this fashion
13 16
= 4
6 +
5 6
2
+ 1
6
3
+ 3
6
4
= 0 .4513
6
.
32 Chapter 3
We may simplify this procedure of successive multiplications by recurring to the following display: 6
13 16
4 6
7 8
5 6
1 4
1 6
1 2
3 The third column contains the integral part of the products of the first column and the second column. Each term of the second
column from the second on is the fractional part of the product obtained in the preceding row. Thus 6 ·
13 16
− 4 = 7
8 , 6
· 7
8 − 5 =
1 4
, etc..
230 Example
Prove that 4 .41
r
is a perfect square in any scale of notation. Solution:
4 .41
r
= 4 + 4
r +
4 r
2
=
2 + 1
r
2
231 Example AIME 1986 The increasing sequence
1 , 3, 4, 9, 10, 12, 13, . . .
consists of all those positive integers which are powers of 3 or sums of distinct powers or 3. Find the hundredth term of the sequence.
Solution: If the terms of the sequence are written in base-three, they comprise the positive integers which do not contain the digit 2. Thus the terms of the sequence in ascending order are
1
3
, 10
3
, 11
3
, 100
3
, 101
3
, 110
3
, 111
3
, . . . In the binary scale these numbers are, of course, the ascending natural numbers 1
, 2, 3, 4, . . .. Therefore to obtain the 100th term of the sequence we write 100 in binary and then translate this into ternary: 100 = 1100100
2
and 1100100
3
= 3
6
+ 3
5
+ 3
2
= 981 .
232 Example AHSME 1993 Given 0
≤ x 1, let
x
n
=
8 :
2x
n−1
if 2x
n−1
1, 2x
n−1
− 1 if
2x
n−1
≥ 1. for all integers n
0. For how many x is it true that x
= x
5
? Solution: Write x
in binary, x
=
∞
X
k=1
a
k
2
k
, a
k
= 0 or 1 .
The algorithm given moves the binary point one unit to the right. For x to equal x
5
we need 0 .a
1
a
2
a
3
a
4
a
5
a
6
a
7
. . .
2
= .a
6
a
7
a
8
a
9
a
10
a
11
a
12
. . .
2
. This will happen if and only if x has a repeating expansion with a
1
a
2
a
3
a
4
a
5
as the repeating block. There are 2
5
= 32 such blocks. But if a
1
= a
2
= ··· = a
5
= 1 then x = 1, which lies outside ]0
, 1[. The total number of values for which x
= x
5
is therefore 32 − 1 = 31 .
Practice
Well-Ordering Principle 33
