## BdMO National 2012: Primary 5

Zzzz
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### BdMO National 2012: Primary 5

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?
Every logical solution to a problem has its own beauty.

prantick
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### Re: BdMO National 2012: Primary 5

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)

ahsaf
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Joined: Mon Jan 19, 2015 10:48 pm

### Re: BdMO National 2012: Primary 5

i am confused Men are born with the reason to help others.
But I realised they strive to become famous.  tanmoy
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### Re: BdMO National 2012: Primary 5

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 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\$.
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