three consecutive integers are as such when they are taken in increasing order and multiplied by 2,3,4 and respectively they add up to 56. find these number
Answers
Answer:
See a solution process below:
Explanation:
First, let's name the three consecutive integers.
Let's call the first integer:
n
Then the next two integers will be
(
n
+
1
)
and
(
n
+
2
)
If we then multiply them as described in the problem and sum these products to 56 we can write an equation as:
2
n
+
3
(
n
+
1
)
+
4
(
n
+
2
)
=
56
We can now solve this equation for
n
:
2
n
+
(
3
×
n
)
+
(
3
×
1
)
+
(
4
×
n
)
+
(
4
×
2
)
=
56
2
n
+
3
n
+
3
+
4
n
+
8
=
56
2
n
+
3
n
+
4
n
+
3
+
8
=
56
(
2
+
3
+
4
)
n
+
(
3
+
8
)
=
56
9
n
+
11
=
56
9
n
+
11
−
11
=
56
−
11
9
n
+
0
=
45
9
n
=
45
9
n
9
=
45
9
9
n
9
=
5
n
=
5
Therefore:
n
+
1
=
5
+
1
=
6
n
+
2
=
5
+
2
=
7
The three consecutive integers are: 5, 6, 7
Step-by-step explanation:
First, let's name the three consecutive integers.
Let's call the first integer:
n
Then the next two integers will be
(
n
+
1
)
and
(
n
+
2
)
If we then multiply them as described in the problem and sum these products to 56 we can write an equation as:
2
n
+
3
(
n
+
1
)
+
4
(
n
+
2
)
=
56
We can now solve this equation for
n
:
2
n
+
(
3
×
n
)
+
(
3
×
1
)
+
(
4
×
n
)
+
(
4
×
2
)
=
56
2
n
+
3
n
+
3
+
4
n
+
8
=
56
2
n
+
3
n
+
4
n
+
3
+
8
=
56
(
2
+
3
+
4
)
n
+
(
3
+
8
)
=
56
9
n
+
11
=
56
9
n
+
11
−
11
=
56
−
11
9
n
+
0
=
45
9
n
=
45
9
n
9
=
45
9