Question

Find the number of solutions to the equation
1a+1b+1c+1d=1
where a, b, c, d are positive integers and a≤b≤c≤d.

Answer 1

sorry here 1a+1b+1c is actually 1over a ... and for b and c also

Answer 2

pls reply

Answer 3

so what is the quesiton? 1/a+1/b+1/c+1/d=1?

Answer 4

@Khushal_Shah I think the question is pretty clear in here ;) but yes, the equation is not clear. If a≤b≤c≤d and they're all positive, a + b + c + d won't be equal to 1.

Answer 5

@tyteen4a03 i was referring to @abhishekdon 's reply to his original question where he says something odd. He never specified anything about d, so i just want to know what the actual question is, lol

Answer 6

the question will be\[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=1\]with\[a\le b\le c\le d\]solution will be a little long, but here is the start point

suppose \(a\ge5\) then\[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d} \le \frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}=0.8<1 \] so we must have \(a<5\) and clearly \(a\) can not be \(1\) so the only values for \(a\) are \(2,3,4\)

Answer 7

now let \(a=4\) the equation becomes\[\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=\frac{3}{4}\]with\[4 \le b\le c \le d\]with a similar reasoning u can realize that \(b=c=d=4\)
only solution for this case is \((a,b,c,d)=(4,4,4,4)\).

Answer 8

let \(a=3\) equation will be\[\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=\frac{2}{3}\approx0.66\]with\[3 \le b\le c \le d\]suppose \(b \ge 5\) then\[\frac{1}{b}+\frac{1}{c}+\frac{1}{d} \le \frac{1}{5}+\frac{1}{5}+\frac{1}{5}=0.6<0.66 \]so we must have \(b<5\) so \(b=3,4\) and....


i hope this helps :)