If x3 + y = 2249, where x and y are natural numbers and HCF of x and y is not 1, then what is the value of (x+y)? Share with your friends Share 2 Neha Sethi answered this Dear student We have, x3+y=2249 Let x=13So x3=133=2197, which is nearer to 2249 y=2249-2197=52So, 133+52=2249 where x=13 and y=52 H.C.F of x=13 and y=52 is 13So x+y=13+52=65 Regards 5 View Full Answer Saurav answered this tiur 1 Gariman answered this x3+y=2249 133+52=2249 (BECAUSE 133=2197) THEREFORE, x=13,y=52 H.C.F. OF x & y i.e. 13 & 52 IS 13. x+y=13+52=65 THEREFORE, 65 IS THE ANSWER. 4 Sai Vignesh Follow Me @sigvins... answered this okko 0