Question 404050: consider the following infinite sequence of numbers 17, 24, 31, 38, 45, 52. what is the least term in the sequence that is a multiple of 2, 3, 4, 5, 6?
Found 2 solutions by Edwin McCravy, richard1234:
Answer by Edwin McCravy(20064)   (Show Source):
You can put this solution on YOUR website!
17, 24, 31, 38, 45, 52
That is an arithmetic sequence with a1 = 17 and d = 7
The nth term of an arithmetic sequence is given by
an = a1 + (n - 1)d
an = 17 + (n - 1)(7)
an = 17 + 7n - 7
an = 10 + 7n
To be a multiple of 2, 3, 4, 5, 6, is to be a multiple of 2*2*3*5 = 60
So we are looking for the smallest integer k so that
10 + 7n = 60k
Write the 10 and the 60 in terms of their nearest multiples of 7.
10 is nearest 7 and 60 is nearest 63
(7 + 3) + 7n = (63 - 3)k
7 + 3 + 7n = 63k - 3k
Divide through by 7
1 +  + n = 9k - k
Isolate the fractional terms:
+ k = 9k - n - 1
Since the right side is an integer, the left side is also an integer.
Let that integer be A.
+ k = A and 9k - n - 1 = A
Clear of fractions:
3 + 3k = 7A
Write the 7 in terms of its nearest multiple of 3, which is 6
3 + 3k = (6 + 1)A
3 + 3k = 6A + A
Divide through by 3
1 + k = 2A + 
Isolate the fraction:
1 + k - 2A =
Since the left side is an integer, the right side is also an integer.
Let that integer be B.
= B and 1 + k - 2A = B
Clear of fractions:
A = 3B
Substituting in 1 + k - 2A = B
1 + k - 2(3B) = B
1 + k = 6B + B
1 + k = 7B
k = 7B - 1
Substituting for A and k in 9k - n - 1 = A
9(7B - 1) - n - 1 = 3B
63B - 9 - n - 1 = 3B
63B - 10 - n = 3B
60B - 10 = n
The smallest possible value of B that would produce
a positive value for n is 1, thus B = 1
60(1) - 10 = n
50 = n
and k = 7B - 1 = 7(1) - 1 = 6
Therefore the desired term is the 50th. That term can be
gotten from either
10 + 7n = 10 + 7(50) = 10 + 350 = 360
or from
60k = 60(6) = 360.
So the answer is 360, which is the 50th term.
Edwin
Answer by richard1234(7193)  (Show Source):
You can put this solution on YOUR website! We want the least term that is a multiple of the LCM of 2, 3, ..., 6, or 60.
Our sequence is denoted by 17 + 7k, where k is an integer. Writing everything modulo 2, 3, and 5, we obtain
1 + k ≡ 0 (mod 2) --> k ≡ 1 (mod 2)
2 + k ≡ 0 (mod 3) --> k ≡ 1 (mod 3)
2 + 2k ≡ 0 (mod 5) --> 1 + k ≡ 0 (mod 5) --> k ≡ 4 (mod 5)
The first two equations imply k ≡ 1 (mod 6), so k can be 1, 7, 13, 19, 25, 31, ... The first value of k that is 4 modulo 5 is 19, so the smallest term in the sequence that satisfies is 17 + 7(19), or 150.
|
|
|
