Download (1) Use the Euclidean algorithm to find the highest common factor of

MATH 115A SOLUTION SET II
JANUARY 20, 2005
(1) Use the Euclidean algorithm to find the highest common factor of 18564 and 30030.
Check your answer by writing each number as the product of prime powers.
Solution:
Applying the Euclidean algorithm yields the following:
30030 = 18564 × 1 + 11466
18564 = 11466 × 1 + 7098
11466 = 7098 × 1 + 4368
7098 = 4368 × 1 + 2730
4368 = 2730 × 1 + 1638
2730 = 1638 × 1 + 1092
1638 = 1092 × 1 + 546
1092 = 546 × 2.
It follows that the highest common factor of 18564 and 30030 is 546.
The prime factorisations of 18564 and 30030 are
18564 = 22 × 3 × 7 × 13 × 17,
30030 = 2 × 3 × 5 × 7 × 11 × 13.
It follows from these prime factorisations that the HCF of the two numbers is
2 × 3 × 7 × 13 = 546.
(2) Which of the following Diophantine equations cannot be solved? (You should justify
your answers.)
(a) 6x + 51y = 22.
(b) 33x + 14y = 115.
(c) 14x + 35y = 93
Solution:
(a) (6, 51) = 3 and 3 - 22. Hence this equation is not soluble.
(b) (33, 14) = 1, and 1 | 115. hence this equation is soluble.
(c) (14, 35) = 7, and 7 - 93. Hence this equation is not soluble.
(3) Find a solution to the Diophantine equation 172x + 20y = 1000.
Solution:
1
2
MATH 115A SOLUTION SET II JANUARY 20, 2005
Applying the Euclidean algorithm to evaluate (172, 20), we find that
172 = 8 × 20 + 12
20 = 1 × 12 + 8
12 = 1 × 8 + 4
8 = 2 × 4,
and so we have that (172, 20) = 4. Since 4 | 1000, the equation can be solved.
To obtain the integer 4 as a linear combination of 172 and 20, we work backwards through
the previous calculation as follows:
4 = 12 − 8
= 12 − (20 − 12)
= 2 × 12 − 20
= 2(172 − 8 × 20) − 20
= 2 × 172 + (−17)20.
Multiplying this relation by 250 yields
1000 = 250 × 4 = 250[2 × 172 + (−17)20]
= 500 × 172 + (−4250)20.
Hence one solution of the equation is x = 500 and y = −4250.
(4) (Euler, 1770) Divide 100 into two summands such that one is divisible by 7 and the
other by 11.
Solution:
Suppose that x + y = 100, with x/7 = k and y/11 = j; then 7j + 11k = 100. This
equation has the general solution k = −300 + 11t, y = 200 − 7t, with the only positive
solution occuring when t = 28. Then k = 8, j = 4, so that x = 56 and y = 44.
9
9
(5) Find the last two digits of the number 99 . [Hint: 99 ≡ 9 (mod 10); hence 99 = 99+10k ;
now use the fact that 99 ≡ 89 (mod 100).]
Solution:
Since
910 = (10 − 1)10 ≡ 1 (mod 100),
we have that 9 · 99 ≡ −99 (mod 100), which in turn implies that 99 ≡ −11 ≡ 89 (mod 100).
Hence
9
99 = 99+10k
= (99 )(910 )k
≡ 89 · 1
(mod 100),
9
and so the last two digits of the number 99 are 89.
(6) Show that 2n divides an integer N if and only if 2n divides the number made up of
the last n digits of N .
MATH 115A SOLUTION SET II
JANUARY 20, 2005
3
Solution:
Observe that 10k = 2k 5k ≡ 0 (mod 2n ) for k ≥ n. Hence if
N = am 10m + · · · + a1 10 + a0 ,
then
N ≡ an−1 10n−1 + · · · + a1 10 + a0
Therefore N ≡ 0 (mod 2n ) if and only if
(mod 2n ).
an−1 10n−1 + · · · + a1 10 + a0 ≡ 0
(mod 2n ),
i.e. if and only if 2n | an−1 . . . a1 a0 .
(7) (i) Let N = am 10m + . . . a2 102 + a1 10 + a0 , where 0 ≤ ak ≤ 9 be the decimal expansion
of a positive integer N . Prove that 7, 11 and 13 all divide N if and only if 7, 11 and 13
divide the integer
M = (100a2 + 100a1 + a0 ) − (100a5 + 100a4 + a3 ) + (100a8 + 100a7 + a6 ) − . . . .
[Hint: If n is even, then 103n ≡ 1, 103n+1 ≡ 10, 103n+2 ≡ 100 (mod 1001); if n is odd, then
103n ≡ −1, 103n+1 ≡ −10, 103n+2 ≡ −100 (mod 1001).]
(ii) Show that the number 29310478561 is divisible by 7 and 13, but not by 11.
Solution:
(i) Applying the hint, we see that, modulo 1001, we have that
N ≡ (100a2 + 100a1 + a0 ) − (100a5 + 100a4 + a3 ) + (100a8 + 100a7 + a6 ) − . . . .
So N ≡ 0 (mod 1001) if and only if the above expression is congruent to 0 modulo 1001 =
7 · 11 · 13.
(ii) From (i), if 29310478561 is divisible by 7, 11 or 13, then so is
561 − 478 + 310 − 29 = 364.
Since 364 = 47̇ · 13, it therefore follows that 29310478561 is divisible by 7 and 13, but not
by 11.
(8) An old and somewhat illegible invoice shows that 72 canned hams were purchased for
$x67.9y. Find the missing digits.
Solution:
We first observe that 72 = 9 × 8. Since 72 | x679y, it follows that 9 | x679y, and so we
must have x + y ≡ 5 (mod 9) (since the alternating sum of the digits of x679y must be
divisible by 9).
Also, since 72 | x6795, it follows that 8 | x679y, and this in turn implies that 8 | 79y (see
Problem 6 on this sheet). We now deduce that y = 2; consequently, x = 3.