The given numbers are 320 and 457Now as 5 and 7 are remainders on division of 320 and 457 by said numberOn subtracting the reminders 5 and 7 from 320 and 457 respectively we get:320 – 5 = 315,457 – 7 = 450The prime factorization:of 315 and 405 are{tex}315 = 3 \\times 3 \\times 5 \\times 7{/tex}{tex}= 3 ^ { 2 } \\times 5 \\times 7{/tex}{tex}450 = 2 \\times 3 \\times 3 \\times 5 \\times 5{/tex}{tex}= 2 \\times 3 ^ { 2 } \\times 5 ^ { 2 }{/tex}{tex}\\therefore{/tex}\xa0H.C.F. of 315 and 450 =\xa0{tex}3 ^ { 2 } \\times 5 = 9 \\times 5 = 45{/tex}Hence the said\xa0number =45