Find the smallest three digit number that divides 923 and 1229 leaving remainder 5 in each
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The require number when divide 923 and 1229, leaves remainder 5.
Required no.=923-5=918
=1229-5=1224
Long division method attached
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Answered by
1
Answer:
153
Step-by-step explanation:
let, the three digit number be = x
x is the common factor of (923-5)=918 and (1229-5)=1224
The factor of 918 are =1,2,3,6,9,17,18,27,34,51,54,102,153,306,459,918
The factor of 1224 are =1,2,3,4,6,8,9,12,17,18,24,34,36,51,68,72,102,136,153,204,306,408,612,1224,
The common factor of 918 and 1229 are =1, 2, 3, 6, 9, 17, 18, 34, 51, 102, 153 and 306,
therefore the smallest common factor of 918 and 1224 is 153
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