BdMO National 2012: Primary 5

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Zzzz
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BdMO National 2012: Primary 5

Unread post by Zzzz » Sun Feb 12, 2012 1:52 pm

Problem 5:
If a number is multiplied with itself thrice, the resultant is called its cube. For example: $3 × 3 × 3 = 27$, hence $27$ is the cube of $3$. If $1,\ 170$ and $387$ are added with a positive integer, cubes of three consecutive integers are obtained. What are those three consecutive integers?
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prantick
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Re: BdMO National 2012: Primary 5

Unread post by prantick » Sun Jan 11, 2015 11:55 am

answer 171
387 + 170 + 1 = 558
1 cube 1
2 cube 8
3 cube 27
4 cube 64
5 cube 125
6 cube 216
7 cube 343
8 cube 512
9 cube 729

729 - 558 = 171 (ans)

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ahsaf
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Re: BdMO National 2012: Primary 5

Unread post by ahsaf » Fri Jan 30, 2015 3:01 pm

i am confused :?
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tanmoy
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Re: BdMO National 2012: Primary 5

Unread post by tanmoy » Fri Jan 30, 2015 8:18 pm

prantick wrote:answer 171
387 + 170 + 1 = 558
1 cube 1
2 cube 8
3 cube 27
4 cube 64
5 cube 125
6 cube 216
7 cube 343
8 cube 512
9 cube 729
729 - 558 = 171 (ans)
What have you done :o Please read the question carefully.The question asked to find the three consecutive integers.
ahsaf wrote:i am confused :?
Here is the solution:
$1,170$ and $387$ are added with a integer.Suppose the integer is $a$.Let $a+1=x^{3},a+170=(x+1)^{3}$ and $a+387=(x+2)^{3}$
Now,$a+1=x^{3}\Rightarrow a=x^{3}-1.......(1)$.
$a+170=(x+1)^{3}\Rightarrow a=(x+1)^{3}-170.........(2)$.
$a+387=(x+2)^{3}\Rightarrow a=(x+2)^{3}-387...........(3)$.
From (1) and (2) we get,
$x^{3}-1=x^{3}+1+3x^{2}+3x-170\Rightarrow 3x^{2}+3x=168\Rightarrow 6x^{2}+6x=336..........(4)$.
From (3) we get,
$a+387=x^{3}+6x^{2}+12x+8\Rightarrow a=x^{3}+6x^{2}+12x-379$.
From (4) we get,$6x^{2}+6x=336$.
$\therefore a=x^{3}-43+6x$........(5)
From (1) and (5) we get,
$x^{3}-1=x^{3}-43+6x\Rightarrow x=7$.
$\therefore$ The digits are $7,8$ and $9$.
"Questions we can't answer are far better than answers we can't question"

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