In how many places did the japanese troops land on december 8, 1941?
1 answer:
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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
Answer:
None
Step-by-step explanation:
By using the equation a^2 +b^2=c^2 we can see if it will work.
Substitute a for 6 and b for 2 to get 6^2+2^2=40
when you you don't get 7 therefore you cant make any triangles