Find the greatest possible no. which can divide 76,132 and 160 and leaving the same remainder in each case.
Answers
Answered by
94
Suppose the greatest number which can divide 76,132 and 160 is p and reminder be r
Also the quotient be q1,q2 and q3 for 76,132 and 160 respectively
Thus
pq1+r=76. (1)
pq2+r=132. (2)
pq3+r=160. (3)
From 1,2 and 3 we get
P(q2-q1)=132-76=56
P(q3-q2)=160-132=28
P(q3-q1)=160-76=84
Therefore the hcf of 56,28 and 84 are
56=2*2*2*7
28=2*2*7
84=2*2*3*7
HCF=2*2*7=28
Therefore the greatest which can divide 76,132 and 160 and leave the same reminder is 28
Answered by
13
Answer:
Step-by-step explanation:
Similar questions