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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.
Answered by Yasmeen Khan | 29 May, 2019, 05:45: PM

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