Using edl find hcf of 506 and 155
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Step-by-step explanation:
a=bq+r
a=506,b=155
506=155*465+41
155=41*3+32
41=32*1+9
32=9*3+5
9=5*1+4
5=4*1+1
4=1*4+0
Therefore H.C.F. Of 506 and 155 is 1
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