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ICSE Class 6 Answered

Find the greatest number that divedes 4410, 5040 and 4725 exactly without leaving any remainder.

Asked by shreeganpationlinehub | 29 May, 2019, 03:48: PM
Expert Answer
Prime factorisation of 4410, 5040 and 4725.
 
4410 = × 3 × 3 × 5 × 7 × 7 
5040 = 2 × × × × 3 × 3 × 5 × 7 
4725 = 3 × 3 × 3 × 5 × × 7
 
HCF(4410, 5040 ,4725) = 3 × 3 × 5 × 7  = 315
 
Therefore the greatest number that divedes 4410, 5040 and 4725 exactly without leaving any remainder is 315.
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