$a(a+b)(a+b+c)(a+b+c+d)=783$

Given that $a,b,c,d$ are positive integers and that they satisfy the above equation, find their product: $a \times b \times c \times d$ .

This is part of the set Trevor's Ten

Based off of one of Sualeh Asif's posts.

The answer is 240.

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I agree with what @Prasun Biswas has to say and I also have a better solution to this problem .

Look at the pic below , the Brilliant Team helped me by giving the answer :P

A Former Brilliant Member
- 6 years, 3 months ago

The key point that one needs to note is that since $a$ is a factor of the LHS expression, it must be a factor of $783$ . Hence, this motivates us to observe the prime factorization of $783$ .

Prasun Biswas
- 6 years, 3 months ago

Great adaption!!

Sualeh Asif
- 6 years, 3 months ago

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Thanks Suelah, yours was much better however. I wanted to make one easier than the rest problem because this set only had hard problems lol.

Trevor Arashiro
- 6 years, 3 months ago

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Well i am trying to mend that problem. In the first one I underestimated the power of the variables. The mended version is coming soon

Sualeh Asif
- 6 years, 3 months ago

Equivalently, letting $a_1=a$ $a_2=a+b$ $a_3=a+b+c$ $a_4=a+b+c+d$ means that $a_1,a_2,a_3,a_4$ is a strictly increasing positive integer sequence such that $a_1a_2a_3a_4=783$

Daniel Liu
- 6 years, 3 months ago

Since you "cheated", I'll cheat too!

(haha nope, I really did not cheat)

Justin Quintos
- 6 years, 2 months ago

```
for (int i=0;i<100;i++)
{
for (int j=0;j<100;j++)
{
for (int k=0;k<100;k++)
{
for (int l=0;l<100;l++)
{
if (i*(i+j)*(i+j+k)*(i+j+k+l)==783)
{
if (i!=0 && j!=0 && k!=0 && l!=0)
{
cout<<i<<" "<<j<<" "<<k<<" "<<l<<endl;
cout<<i*j*k*l<< endl;
break;
}
}
}
}
}
}
```

}

Zeeshan Ali
- 6 years, 2 months ago

(a + b + c + d) > (a + b + c) > (a + b) > a

783 = 1 X 3 X 9 X 29

a = 1

a + b = 3

a + b + c = 27

a + b + c + d = 29

a = 1 , b = 2 , c = 6 , d = 20

Then

a X b X c X d = 240

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I agre with trevor arashiro

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I did same as Trevor Arashiro

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Since a,b,c,d are all positive, $(a+b+c+d)>(a+b+c)>(a+b)>a$

If we factor 783 we get

$1\cdot 3 \cdot 9\cdot 29$

$3\cdot 3 \cdot 3 \cdot 29$

$1\cdot 1 \cdot 27 \cdot 29$

Since all are positive, $a+b+c+d\neq a+b+c \neq a+b \neq a$ , we can eliminate the second and third possibilities.

Thus we have

$\begin{aligned} 1&=a\\ 3&=a+b\\ 9&=a+b+c\\ 29&=a+b+c+d\\ \end{aligned}$

Thus some simple algebra yields $a=1,~b=2,~c=6,~d=20$

$\therefore abcd=1\cdot2\cdot6\cdot20=\boxed{240}$