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x is the smallest positive integer such that when it is divided by 7, 8 and 9 leaves remainder as 4, 5 and 6 respectively. Find the remainder when x3 + 2x2 – x – 3 is divided by 132.
(2015)
  • a)
    49
  • b)
    76
  • c)
    94
  • d)
    15
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
x is the smallest positive integer such that when it is divided by 7, ...
x = L.C.M. of (7, 8, 9) – 3 = 504 – 3 = 501
x3 + 2x2 + x – 3 = (x – 1) (x + 1) (x + 2) –1
= 500 × 502 × 503 – 1
Remainder when 500 × 502 × 503 – 1 is divided by:
11 = 4
3 = 0
4 = 3
Required remainder = least possible number which when divided by 11, 3 and 4 leavs remainder 4, 0 and 3 respectively.
Such least no. is 15.
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Most Upvoted Answer
x is the smallest positive integer such that when it is divided by 7, ...
To find the remainder when $x^{3}-2x^{2}$ is divided by $7,8,$ and $9,$ we can find the remainders of $x^{3}$ and $2x^{2}$ when divided by $7,8,$ and $9,$ and then subtract.

$x\equiv4\pmod7$ implies $x^{2}\equiv2\pmod7$ and $x^{3}\equiv8\equiv1\pmod7.$

$x\equiv5\pmod8$ implies $x^{2}\equiv1\pmod8$ and $x^{3}\equiv5\pmod8.$

$x\equiv6\pmod9$ implies $x^{2}\equiv0\pmod9$ and $x^{3}\equiv0\pmod9.$

Thus, $x^{3}-2x^{2}\equiv1-2(5)\equiv\boxed{3}\pmod8.$
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x is the smallest positive integer such that when it is divided by 7, 8 and 9 leaves remainder as 4, 5 and 6 respectively. Find the remainder when x3 + 2x2 – x – 3 is divided by 132.(2015)a)49b)76c)94d)15Correct answer is option 'D'. Can you explain this answer?
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