Question

if d is the hcf of 160 and 24 then find x and y satisfying d =160x plus 24y. also show that x and y are not unique

Answer 1

$Givendisthehcfof160and24andwehavetofindthevalueofxandysuchthat$

$d=160x+24y$

$Alsohavetoshowthattheyarenotunique$

$NowbyEucli{d}^{\prime }sDivisionLemma,HCFdof160and24is$

$160=24\times 6+16.......\left(i\right)$

$24=16\times 1+8........\left(ii\right)$

$16=8\times 2+0$

$henceremainder=0$

$thereforeHCFd=8$

$Now8=24-16\times 1.....\left\{from\left(i\right)\right\}$

$8=24-(160-24\times 6)\times 1.....\left\{from\left(ii\right)\right\}$

$8=24-160+24\times 6$

$8=160\times (-1)+24\times 7......\left(a\right)$

$\therefore x=-1,y=7$

$Nowaddingandsubtracting160\times 24ineqn\left(a\right),weget$

$8=160\times (-1)+24\times 7+160\times 24-160\times 24$

$8=160\times (-1+24)+24(7-160)$

$8=160\times 23+24(-153)$

$fromtheaboveweitisclearthatvalueofxandyisnotuniques$