Finding Integer Solutions Worksheet With Answer Key - 7th Grade, Serin Hong Page 3
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6Math 7 Spring 2016
Instructor: Serin Hong
3. (10 pts) Find all the integer solutions of 6x + 15y + 10z = 8.
Solution. Since gcd(6, 15) = 3, we can write 6x + 15y = 3q where q is an integer.
Now the equation becomes 3q + 10z = 8. We easily find that 3 ( 3) + 10 1 = 1, and multiplying both
sides by 8 yields 3
( 24) + 10 8 = 8. Then we can solve the equation 3q + 10z = 8 in integers by
q =
24
10t,
z = 8 + 3t where t
(1)
Z.
Next we solve the equation 6x + 15y = 3q. Since 6 ( 2) + 15 1 = 3, multiplying both sides by q gives
6
15
6 ( 2q) + 15 q = 3q. Since
= 2 and
= 5, we parametrize all the integer solutions by
gcd(6, 15)
gcd(6, 15)
x =
2q
5s,
y = q + 2s where s
(2)
Z.
Now substituiting (1) to (2) yields
x = 48 + 20t
5s,
y =
24
10t + 2s,
z = 8 + 3t where t, s
Z.
4. (15 pts) Determine the number of
integer solutions of 2x + 3y = 300.
Solution. Let us first solve the equation 2x + 3y = 300 in integers. From 2 ( 1) + 3 1 = 1 we obtain
2 ( 300) + 3 300 = 300. Since gcd(2, 3) = 1, the integer solutions are given by
x =
300
3t,
y = 300 + 2t where t
Z.
Now x > 0 if and only if
300 3t > 0, or equivalently t <
100. Similarly, y > 0 if and only if 300 + 2t > 0,
or equivalently t >
150. Hence x and y are positive integers if and only if
150 < t <
100. This gives 49
possible values for t, namely
101, 102,
, 149. Hence the number of positive integer solutions is 49.
5. Recall that the Fibonacci sequence (F )
is defined by the recurrence relation F
= F
+ F for
1
+2
+1
n
1 with initial values F
= 1, F
= 1.
1
2
(a) (10 pts) Find gcd(F
, F ).
+2
Solution. For n = 1, we directly compute gcd(F
, F ) = gcd(F
, F
) = 1.
+2
3
1
Let us now assume that n
2. From the recurrence relation we find
F
= F
+ F = (F + F
) + F = 2F + F
.
+2
+1
1
1
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