Show that 24n-1 is
divisible by 15, if n be a positive integer.Show that 72m I, where
is any positive integer, is divisible by each 2,3,4,6,8,12,16, 24,48.
Show that am(a-1)+bm(b-1)
is not divisible by a+b, m being any positive integer
Show that xn-nx+n-1 is exactly divisible
(x-1)2
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Assuming the question as 24n 1 is divisible by 15
This problem solving needs the knowledge of mathematical induction.
Let p(n): 24n 1 is divisible by 15
Put n = 1 in p(n)
p(1) = 24n 1 = 16 1 =15
Hence p(n) is divisible by 15 for n=1 → (1)
Now let us assume that p(n) is true for n=2,3,4n → (2)
That is 24n 1 = 15k where k is any integer
⇒ 24n =15k+1 → (3)
We should prove that p(n) is also true for (n+1)
Consider p(n+1): 24(n + 1) 1
= 24n × 24 1
= (15k+1) × 24 1 [From equation (3)]
=15k × 16 + 16 1
= 15k × 16 + 15
= 15(16k + 1)
Clearly, p(n+1) is divisible by 15 →(4)
From (1), (2) and (4) by principle of mathematical induction, we have 24n 1 is divisible by 15 where n is any positive integer.
Hope This Helps :)
This problem solving needs the knowledge of mathematical induction.
Let p(n): 24n 1 is divisible by 15
Put n = 1 in p(n)
p(1) = 24n 1 = 16 1 =15
Hence p(n) is divisible by 15 for n=1 → (1)
Now let us assume that p(n) is true for n=2,3,4n → (2)
That is 24n 1 = 15k where k is any integer
⇒ 24n =15k+1 → (3)
We should prove that p(n) is also true for (n+1)
Consider p(n+1): 24(n + 1) 1
= 24n × 24 1
= (15k+1) × 24 1 [From equation (3)]
=15k × 16 + 16 1
= 15k × 16 + 15
= 15(16k + 1)
Clearly, p(n+1) is divisible by 15 →(4)
From (1), (2) and (4) by principle of mathematical induction, we have 24n 1 is divisible by 15 where n is any positive integer.
Hope This Helps :)
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3
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