783 Followers Problem
a ( a + b ) ( a + b + c ) ( a + b + c + d ) = 7 8 3
Given that a , b , c , d are positive integers and that they satisfy the above equation, find their product: a × b × c × 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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6 solutions
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
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 7 8 3 . Hence, this motivates us to observe the prime factorization of 7 8 3 .
Great adaption!!
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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.
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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
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 1 a 2 a 3 a 4 = 7 8 3
Since you "cheated", I'll cheat too!
i really did not cheat on this problem tho
(haha nope, I really did not cheat)
include<iostream>
using namespace std; int main() {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;
}
}
}
}
}
}
}
(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
I agre with trevor arashiro
I did same as Trevor Arashiro
S i n c e 7 8 3 = 1 ∗ 3 ∗ 9 ∗ 2 9 , a n d e a c h p a r e n t h e s e s h a s a v a l u e i n c r e a s i n g f r o m l e f t t o r i g h t . T h u s a = 1 , . . . . . . . ( a + b ) = 1 + b = 3 , s o b = 2 , . . . . . . ( a + b + c ) = 3 + c = 9 , s o c = 6 , . . . . . . ( a + b + c + d ) = 9 + d = 2 9 , s o d = 2 0 . . . . . . . . a ∗ b ∗ c ∗ d = 1 ∗ 2 ∗ 6 ∗ 2 0 + 2 4 0 .
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 ⋅ 3 ⋅ 9 ⋅ 2 9
3 ⋅ 3 ⋅ 3 ⋅ 2 9
1 ⋅ 1 ⋅ 2 7 ⋅ 2 9
Since all are positive, a + b + c + d = a + b + c = a + b = a , we can eliminate the second and third possibilities.
Thus we have
1 3 9 2 9 = a = a + b = a + b + c = a + b + c + d
Thus some simple algebra yields a = 1 , b = 2 , c = 6 , d = 2 0
∴ a b c d = 1 ⋅ 2 ⋅ 6 ⋅ 2 0 = 2 4 0