Question 1011506


{{{n}}} and {{{n+2}}} the sum of the reciprocals of the integers is {{{29/420}}}

for integer {{{n}}} reciprocal is {{{1/n}}}
for integer {{{n+2}}}  reciprocal is {{{1/(n+2)}}}

if the sum of the reciprocals  is {{{29/420}}}, we have

{{{1/n+1/(n+2)=29/420}}}........solve for {{{n}}}

{{{1(n+2)/(n(n+2))+(1*n)/(n(n+2))=29/420}}}

{{{(n+2+n)/(n(n+2))=29/420}}}

{{{(2n+2)/(n^2+2n)=29/420}}}...cross multiply

{{{420(2n+2)=29(n^2+2n)}}}

{{{840n+840=29n^2+58n}}}

{{{0=29n^2+58n-840n-840}}}

{{{29n^2-782n-840=0}}}...factor

{{{29n^2-812n+30n-840=0}}}

{{{(29n^2-812n)+(30n-840)=0}}}

{{{29n(n-28)+30(n-28)=0}}}

{{{(n-28) (29 n+30) = 0}}}

solutions:

if {{{(n-28)  = 0}}}=>{{{n=28}}}
if {{{ (29 n+30) = 0}}}=>{{{n=-30/29}}}-> since we need an integer, disregard this solution

so, your integers are: {{{n=28}}} and {{{n+2=30}}}
 their reciprocals are: {{{1/28}}} and {{{1/30}}}

check their sum: {{{1/28+1/30}}} ....since {{{LCM (28,30)=420}}}, we have
{{{15/420+14/420=29/420}}}