Class X Ncert apply euclids division algonthm to find hcf of no. 4052 and 420
Answers
Answered by
2
the HCF of 4052 and 420
euclid division lemma
a=bq+r
4052 = 420(9)+272
420=272(1)+148
272=148(1)+124
148=124(1)+24
124=24(5)+4
24=4(5)+0
so HCF is 4
euclid division lemma
a=bq+r
4052 = 420(9)+272
420=272(1)+148
272=148(1)+124
148=124(1)+24
124=24(5)+4
24=4(5)+0
so HCF is 4
Answered by
1
Euclids Division Lemma: a=bq+r (0≤r<b)
4052=420*9+272
420=272*1+148
272=148*1+124
148=124*1+24
124=24*5+4
24 =4*6+0
∴HCF(4052,420)=4
4052=420*9+272
420=272*1+148
272=148*1+124
148=124*1+24
124=24*5+4
24 =4*6+0
∴HCF(4052,420)=4
Similar questions