n=2p-3 , p ∈ N and divide n² by 4 what will be the remainder?
Answers
Given : n =2p-3, p ∈ N ,
To Find : remainder if divide n² by 4
Solution:
n =2p-3,
Squaring both sides
=> n² = (2p - 3)²
=> n² = (2p)² - 2(2p)3 + 3²
=> n² = 4p² - 4p(3) + 9
=> n² = 4.p² - 4.3p + 8 + 1
=> n² = 4(p² - 3p + 2 ) + 1
=> n² = 4k + 1
if divide n² by 4 then remainder = 1
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Given : n = 2p - 3, p ∈ N and divide n² by 4. what will be the remainder ?
Solution :
→ n = (2p - 3)
→ n² = (2p - 3)² = (2p)² + (3)² - 2*2p*3 = 4p² - 12p + 9.
now , given that, p ∈ N and n² is divided by 4,
→ (n²/4)
→ (4p² -12p +9)/4
→ (4p²/4) - (12p/4) + (9/4)
as we can see,
- 4p² is a factor of 4 for p ∈ N . => Remainder 0.
- 12p = (4 * 3p) is a factor of 4 for p ∈ N => Remainder O.
- 9/4 = Remainder 1.
- 0 - 0 + 1 = 1
Therefore, we can conclude that, for p ∈ N, when n² is divided by 4, remainder will be 4.
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