Find the HCF, by Euclid’s division algorithm of the numbers 92695, 7384 and 7168 leaving remainders 5, 6 and 7 respectively
Answers
HCF of 92695, 7384 and 7168 is 7 leaving remainders 5, 6 & 7
Step by step explanation:
Since the numbers, 92695, 7385 &7168 leave the remainders 5, 6 & 7 respectively, thus, we subtract the remainders from the numbers,
therefore,
92695-5=92690
7384-6=7378
7168-7=7161
therefore, the new formed are,
92690, 7378 and 7161
since 92690 is greater than 7378, thus we take 92690 as 'a', and 7378 as 'b'
therefore, by Euclid's division algorithm,
a=bq+c
92690= 7378x12+4154
7378= 4154x1+3220
4154= 3220x1+930
3220= 930x3+430
930=430x2+70
430=70x6+10
70=10x7+0
therefore, HCF of 92690 & 7378 is 7 ...(1)
since 7161 is greater than 7, we take 'a' as 7161 and 'b' as 7,
therefore,
7161=7x1023+0
therefore, HCF of 7161 & 7 is 7 ...(2)
therefore, HCF of 92690, 7378 & 7161 is 7...
therefore, from (1) & (2)
HCF of 92695, 7384 and 7168 is 7 leaving remainders 5, 6 & 7
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