Divisibility
Igor Zelenko
SEE Math,
August 13-14, 2012
Before getting started
Below is the list of problems and games I prepared for our activity.
We will certainly solve/discuss/play only part of them and maybe
in different order. It is recommended to continue to work at home
on the problem we will not have time to discuss today.
Problem 1
Problem 1
In one kingdom there are only two kinds of coins: a coin of 9 cents
and a coin of 15 cents. Can we get 70 cents using these coins?
Remark
In general, given integer numbers a, b, and c, if the equation
ax + by = c
has an integer solutions x and y , then c must be divisible by the
greatest common divisor (gcd) of a and b.
Remark
In general, given integer numbers a, b, and c, if the equation
ax + by = c
has an integer solutions x and y , then c must be divisible by the
greatest common divisor (gcd) of a and b.
Moreover (for further thinking), if c is divisible by gcd (a,b) then
the equation ax + by = c has integer solutions (in fact infinitely
many). The latter is based on the so-called Euclid’s algorithm for
finding gcd.
Problem for home
Three piles of stones contain 51, 49 and 5 stones respectively. It is
allowed to unify any two piles to one pile and to divide a pile
containing an even number of stones into two piles with the equal
number of stones. Is it possible to get in this way 105 piles of one
stone.
Problem 2
Problem 2
(a) Show that a square of a natural number has an odd number
of divisors.
Problem 2
(a) Show that a square of a natural number has an odd number
of divisors.
(b) Is the converse statement true,
Problem 2
(a) Show that a square of a natural number has an odd number
of divisors.
(b) Is the converse statement true, i.e., if a number has an odd
number of divisors, then is it a perfect square?
Problem 3
In some kingdom there was a jail with 100 cells, enumerated from
1 to 100. In each cell there was one prisoner. The locks in the cells
were arrange such that the locked door becomes unlocked after
one turn of the key clockwise and the unlocked door became
locked after one turn of the key clockwise.
Problem 3
In some kingdom there was a jail with 100 cells, enumerated from
1 to 100. In each cell there was one prisoner. The locks in the cells
were arrange such that the locked door becomes unlocked after
one turn of the key clockwise and the unlocked door became
locked after one turn of the key clockwise.
The king was in a battle against one of his neighbor and at some
moment he thought he is going to win. Being exited, he sent a
messenger with an order to unlock all cells in the prison. However,
the situation in the battlefield has changed quickly and the king
sent another messenger (in pursuit of the first one) with the order
to turn a key clockwise in each second cell. Then the situation in
the battlefield improved and he sent the third messenger with an
order to turn a key clockwise in each third cell and so on.
Problem 3
In some kingdom there was a jail with 100 cells, enumerated from
1 to 100. In each cell there was one prisoner. The locks in the cells
were arrange such that the locked door becomes unlocked after
one turn of the key clockwise and the unlocked door became
locked after one turn of the key clockwise.
The king was in a battle against one of his neighbor and at some
moment he thought he is going to win. Being exited, he sent a
messenger with an order to unlock all cells in the prison. However,
the situation in the battlefield has changed quickly and the king
sent another messenger (in pursuit of the first one) with the order
to turn a key clockwise in each second cell. Then the situation in
the battlefield improved and he sent the third messenger with an
order to turn a key clockwise in each third cell and so on.
In this way 100 messengers arrived to the jail almost
simultaneously and turned successively the keys according to the
king orders. After this procedure, how many prisoners were free
and from what cells are they?
Divisibility by 9 test, Problem 4
Divisibility by 9 test, Problem 4
(a) Show that the difference of a natural number and the sum of
its digit is divisible by 9.
Divisibility by 9 test, Problem 4
(a) Show that the difference of a natural number and the sum of
its digit is divisible by 9.
(b) Deduce from here the following generalization of the
divisibility by 9 test: the remainders after division by 9 of a
number and the sum of its digits are equal.
Divisibility by 9 test, Problem 4
(a) Show that the difference of a natural number and the sum of
its digit is divisible by 9.
(b) Deduce from here the following generalization of the
divisibility by 9 test: the remainders after division by 9 of a
number and the sum of its digits are equal.
In particular, this implies the very well know divisibility by 9
test: the number (an−1 an−2 . . . a1 a0 )10 is divisible by 9 if and
only of a0 + a1 + . . . + an−1 is divisible by 9.
Applications of divisibility by 9 test: Problem 5
John found the number
100! = 1 · 2 · . . . 100.
Then he sum up all digits of this number, then he sum up all
digits in this new number and so on till he got a one-digit number.
What is this number?
Applications of divisibility by 9 test: Problem 6
After changing the order of digits in some number one gets a
number that is three times bigger than the original one. Prove that
the resulting number is a multiple of 27.
Application of divisibility by 9 test: Problem 7
Find the minimal natural number containing only digits 1 and 0 (in
its decimal form) which is divided by 225.
Divisibility by 11 tests: Problem 8
(a) Show that the number (an−1 an−2 . . . a1 a0 )10 is divisible by 11
if and only of a0 − a1 + a2 − a3 . . . + (−1)n−1 an−1 is divisible
by 11
Divisibility by 11 tests: Problem 8
(a) Show that the number (an−1 an−2 . . . a1 a0 )10 is divisible by 11
if and only of a0 − a1 + a2 − a3 . . . + (−1)n−1 an−1 is divisible
by 11 or, equivalently, the difference of the sum of all odd
digits and the sum of all even digits is divisible by 11.
Divisibility by 11 tests: Problem 8
(a) Show that the number (an−1 an−2 . . . a1 a0 )10 is divisible by 11
if and only of a0 − a1 + a2 − a3 . . . + (−1)n−1 an−1 is divisible
by 11 or, equivalently, the difference of the sum of all odd
digits and the sum of all even digits is divisible by 11.
(b) Using this test, check whether 82918 is divisible by 11.
Divisibility by 7 test: Problem 9
(a) Show that the number (an−1 an−2 . . . a1 a0 )10 is divisible by 7 if
and only if the number the number (an−1 an−2 . . . a1 )10 + 5a0
is divisible by 7.
Divisibility by 7 test: Problem 9
(a) Show that the number (an−1 an−2 . . . a1 a0 )10 is divisible by 7 if
and only if the number the number (an−1 an−2 . . . a1 )10 + 5a0
is divisible by 7.
In other words, we first take the number obtained from the
original one by erasing the last digit (the digit of units) and
then add to it the last digit multiplied by 5.
Divisibility by 7 test: Problem 9
(a) Show that the number (an−1 an−2 . . . a1 a0 )10 is divisible by 7 if
and only if the number the number (an−1 an−2 . . . a1 )10 + 5a0
is divisible by 7.
In other words, we first take the number obtained from the
original one by erasing the last digit (the digit of units) and
then add to it the last digit multiplied by 5.
(b) Using this test, check whether the number 12341 is divisible
by 7.
Divisibility by 7 test: Problem 9
(a) Show that the number (an−1 an−2 . . . a1 a0 )10 is divisible by 7 if
and only if the number the number (an−1 an−2 . . . a1 )10 + 5a0
is divisible by 7.
In other words, we first take the number obtained from the
original one by erasing the last digit (the digit of units) and
then add to it the last digit multiplied by 5.
(b) Using this test, check whether the number 12341 is divisible
by 7.
(c) Deduce a similar test for divisibility by 49.
Divisibility by 13 test: Problem 10
(a) Show that the number (an−1 an−2 . . . a1 a0 )10 is divisible by 13
if and only if the number the number
(an−1 an−2 . . . a1 )10 + 4a0
Divisibility by 13 test: Problem 10
(a) Show that the number (an−1 an−2 . . . a1 a0 )10 is divisible by 13
if and only if the number the number
(an−1 an−2 . . . a1 )10 + 4a0
In other words, we first take the number obtained from the
original one by erasing the last digit (the digit of units) and
then add to it the last digit multiplied by 4.
Divisibility by 13 test: Problem 10
(a) Show that the number (an−1 an−2 . . . a1 a0 )10 is divisible by 13
if and only if the number the number
(an−1 an−2 . . . a1 )10 + 4a0
In other words, we first take the number obtained from the
original one by erasing the last digit (the digit of units) and
then add to it the last digit multiplied by 4.
(b ) Using this test, check whether the number 10205 is divisible
by 13.
Divisibility by 13 test: Problem 10
(a) Show that the number (an−1 an−2 . . . a1 a0 )10 is divisible by 13
if and only if the number the number
(an−1 an−2 . . . a1 )10 + 4a0
In other words, we first take the number obtained from the
original one by erasing the last digit (the digit of units) and
then add to it the last digit multiplied by 4.
(b ) Using this test, check whether the number 10205 is divisible
by 13.
(c) Deduce a similar test for divisibility by 39.
If you ambitious: Problems for home
Find similar test for divisibility by 17, 23, 27, 29, 59, 79, 89 and
more.
Game using divisibility
A pile contains n stones. Two players participate in the game. At
each turn, a player may take no more than p stones from the pile
but he mus take at least one stone. The loser is the player who
cannot move. Which player has the winning strategy (the answer
actually depends on n and p).
Remainders and periodicity. Problem 11
Find the last digit of the number 777777
Remainders and periodicity. Problem 12
Find the remainder when the number 32011 is divided by 7.