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| 1. |
Find the HCF of 4052 and 12576 |
| Answer» 12576 > 4052a=bq+r12576= 4052×3+420 4052=420×9+272 420=272×1+148272= 148×1+124148=124×1+24124=24×5+424=4×6+0HCF =4 | |