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Question: If the remainder of 3<sup>37</sup> is divided by 80 is a and remainder when 4<sup>101</sup> is divid...

If the remainder of 337 is divided by 80 is a and remainder when 4101 is divided by the 101 is b, the quadratic equation whose roots are ab2, ba2 –

A

2x2 – 42x + 1729 = 0

B

x2 – 84x + 1728 = 0

C

3x2 – 82x + 729 = 0

D

none of these

Answer

x2 – 84x + 1728 = 0

Explanation

Solution

337 = 34·9 · 3 = 3 · (81)9 = 3(80 + 1)9

= 3 (9C0 (80)9 + 9C1 (80)8 +……+ 9C9)

Then remainder is 3

Ž a = 3

and 4101 = 4100 · 4

4100 is divisible by 101

Remainder is 4, a2b = 32 · 4 = 36

ba2 = 3 · 42 = 48

equation x2 – 84x + 1728 = 0.