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)?

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  
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  • 5
tiur
  • 1
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
okko
  • 1
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