+816 7. The numbers 19, a, b, c, 73 form an arithmetic sequence, in that order.
Find a+b+c.
8. The number 8!+9!+10! is equal to n*8! for some integer n. What
is the value of n?
9. Regular octagon ABCDEFGH has side length 4 units. What is the
area of square ACEG? Express your answer in simplest radical form.
10. The least common multiple of 210 and n is 210n, where n is an integer
greater than 1. What is the smallest possible value of n?
11. The side lengths of an acute triangle, in inches, are integers and form
an increasing arithmetic sequence. What is the smallest possible value
for the perimeter of the triangle?
Please answer all questions...Thanks!
7)
73 -19 =54
54 / 4 = 13.5 - common difference
19, 32.5, 46, 59.5, 73.........etc.
a + b + c = 32.5 + 46 + 59.5 = 138
8)
[8! + 9! + 10!] =8! x [(1 + 9 + 90)] / 8! x n
n = 100
+36916 9. Interior angles of a regular polyhedron total (n-2)(180)
for an octagon (8-2)(180) = 1080 there are 8 interior angles so each is 1080/8 = 135 degrees
The sides of the SQUARE form isocoles triangles of sides 4, 4 and 'x' (the side of the square) with an included 135 degree angle beteen the 4 and 4 sides
The other two angles are (180-135)/2 = 22.5 degrees
Use the law of sines to find 'x'
sin 135 /x = sin22.5 / 4 results in square side length of 7.39
Square Area = 7.39 x 7.39 =54.63 units^2
+36916 11. The smallest I can find ...by trial and error is 4 5 6
perimeter = 15
+128474 7. The numbers 19, a, b, c, 73 form an arithmetic sequence, in that order.
Find a+b+c.
19 + 4d = 73 subtract 19 from both sides
4d = 54 divide both sides by 4
d = 13.5
Calling m the first term ....the sum of a + b + c is
( m + d ) + (m + 2d) + (m + 3d) =
3m + 6d =
3(19) + 6(13.5) =
57 + 81 =
138
+128474 10. The least common multiple of 210 and n is 210n, where n is an integer
greater than 1. What is the smallest possible value of n?
For 210n to be the LCM, n will be the first positive integer that does not share any factors with 210.....this is 11
LCM 210*11 = 2310
+128474 11. The side lengths of an acute triangle, in inches, are integers and form
an increasing arithmetic sequence. What is the smallest possible value
for the perimeter of the triangle?
Let m be the shortest side, m + d the next longest,and m + 2d the longest
And we have, by the Triangle Inequality :
( m ) + (m + d) > m + 2d
2m + d > m + 2d
m > d
Since we have integers, the smallest possible value of d is 1.....which means the smallest possible value of m must be 2
This means that the the second term, m+ d ≥ 3 and the third term m + 2d ≥ 4
If the triangle is 2,3,4 we must have the following to be acute
4 < √ [ 2^2 + 3^ 2]
4 < √13 which is untrue
A 3,4,5 triangle is right
If the triangle is 4,5,6 we must have that
6 < √ [ 4^2 + 5^2]
6 < √41 which is true
So....4,5,6 is the triangle we want....and the smallest perimeter is 15
