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Solution 3

To prove that 7n - 2n is divisible by 5 for any natural number n, the proof uses mathematical induction. It first shows that if a is divisible by d, then a^2 is also divisible by d. It then assumes the statement is true for some natural number k. It shows that a^k+1 is also divisible by d, following the same pattern. Therefore, the statement 7n ≡ 2n (mod 5) is true for any natural number n, meaning 7n - 2n is divisible by 5.

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Yau Ching Koon
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102 views1 page

Solution 3

To prove that 7n - 2n is divisible by 5 for any natural number n, the proof uses mathematical induction. It first shows that if a is divisible by d, then a^2 is also divisible by d. It then assumes the statement is true for some natural number k. It shows that a^k+1 is also divisible by d, following the same pattern. Therefore, the statement 7n ≡ 2n (mod 5) is true for any natural number n, meaning 7n - 2n is divisible by 5.

Uploaded by

Yau Ching Koon
Copyright
© Attribution Non-Commercial (BY-NC)
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PDF, TXT or read online on Scribd
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To prove 7n 2n is divisible by 5 for n N. Proof: Suppose that a, b, d, q N related by a = dq + b Here d may be known as the divisor and q the quotient.

nt. b may be regarded as the remainder of a when divided by d. This statement is equivalent to a b(mod d). We will seek to prove that an bn (mod d). We will approach the problem by rst principle of induction. Clearly a2 = (dq + b)2 = (dq + b)dq + b2 and so, a2 b2 (mod d). Supposed that the statement is true up to some natural number k, ak+1 = ak a = ak (dq + b) = ak dq + ak b Clearly the rst term is divisible by d and the last gives a remainder of bk as permitted by our assumption.

ak bk (mod 5)
So, we have the corollary 7n 2n (mod 5) or 7n = 2n + 5m

for some integer values of m, which may be written into 7n 2n = 5m This last statement is what we wish to proved.

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