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How many integer solutions are there to a+b+c=1000 and a/5+b/6+c/7=160. Don't use that much calculation. Do not use Ginny, the answer it gives is not right

User Najzero
by
8.0k points

1 Answer

0 votes

Answer:

Step-by-step explanation:Let's solve the system of equations using a systematic approach without resorting to excessive calculation.

Given:

+

+

=

1000

a+b+c=1000

5

+

6

+

7

=

160

5

a

+

6

b

+

7

c

=160

First, from (1), we can express

=

1000

c=1000−a−b.

Now, substituting

c in equation (2), we get:

5

+

6

+

1000

7

=

160

5

a

+

6

b

+

7

1000−a−b

=160.

Multiplying everything by 210 (which is the LCM of 5, 6, and 7) to get rid of the denominators:

42

+

35

+

30

(

1000

)

=

33600

42a+35b+30(1000−a−b)=33600

Expanding and combining like terms:

42

+

35

+

30000

30

30

=

33600

42a+35b+30000−30a−30b=33600

12

+

5

=

3600

12a+5b=3600

12

=

3600

5

12a=3600−5b

=

300

5

12

a=300−

12

5b

For "a" to be an integer,

5

5b must be divisible by 12. The factors of 12 are 1, 2, 3, 4, 6, and 12. Only 4 and 12 can be factors of 5b since 5 is prime. Thus,

=

4

×

3

=

12

b=4×3=12 or

=

12

×

3

=

36

b=12×3=36 are the possible values.

If

=

12

b=12,

=

300

5

(

12

)

12

=

295

a=300−

12

5(12)

=295. From the first equation,

=

1000

295

12

=

693

c=1000−295−12=693.

If

=

36

b=36,

=

300

5

(

36

)

12

=

285

a=300−

12

5(36)

=285. From the first equation,

=

1000

285

36

=

679

c=1000−285−36=679.

Thus, there are two integer solutions to the system of equations:

(a, b, c) = (295, 12, 693) and (285, 36, 679).

User Tishu
by
7.7k points
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