Write the first five terms of a numerical pattern that begins with 5 and then adds 6
1 answer:
This pattern forms an arithmetic sequence with the first term . The difference between two consecutive terms of this sequence is 6. The rule to calculate the other term of the sequence is is to add 6 to the previous term. This can be archived as shown in the calculations below,
The first terms of this sequence are 5, 11, 17, 23,29.
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The equivalence 
means that n-5 is a multiple of 12.
that is
n-5=12k, for some integer k
and so
n=12k+5
for k=-1, n=-12+5=-7
for k= 0, n=0+5=5 (the first positive integer n, is for k=0)
we solve 5000=12k+5 to find the last k
12k=5000-5=4995
k=4995/12=416.25
so check k = 415, 416, 417 to be sure we have the right k:
n=12k+5=12*415+5=4985
n=12k+5=12*416+5=4997
n=12k+5=12*417+5=5009
The last k which produces n<5000 is 416
For all k∈{0, 1, 2, 3, ....416}, n is a positive integer from 1 to 5000,
thus there are 417 integers n satisfying the congruence.
Answer: 417
means that n-5 is a multiple of 12.
that is
n-5=12k, for some integer k
and so
n=12k+5
for k=-1, n=-12+5=-7
for k= 0, n=0+5=5 (the first positive integer n, is for k=0)
we solve 5000=12k+5 to find the last k
12k=5000-5=4995
k=4995/12=416.25
so check k = 415, 416, 417 to be sure we have the right k:
n=12k+5=12*415+5=4985
n=12k+5=12*416+5=4997
n=12k+5=12*417+5=5009
The last k which produces n<5000 is 416
For all k∈{0, 1, 2, 3, ....416}, n is a positive integer from 1 to 5000,
thus there are 417 integers n satisfying the congruence.
Answer: 417