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THE LCM AND GCF OF THREE NUMBERS ARE 360 AND 3. IF TWO OF THE numbers ARE 18 AND 24
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Let the third number be x.
18 = 2·3^2, 24 = 2^3·3, so LCM(18,24) = 2^3·3^2 = 72 and gcd(18,24) = 6.
Given gcd(18,24,x) = 3, we have gcd(6,x) = 3, so x is divisible by 3 but not by 2.
Given LCM(18,24,x) = LCM(72,x) = 360 = 2^3·3^2·5, x must supply the factor 5 but must not introduce any prime power larger than those in 360. Since x is not divisible by 2, its form must be 3^a·5 with a = 1 or 2.
Thus x = 3·5 = 15 or x = 3^2·5 = 45.
Answer: 15 or 45.
18 = 2·3^2, 24 = 2^3·3, so LCM(18,24) = 2^3·3^2 = 72 and gcd(18,24) = 6.
Given gcd(18,24,x) = 3, we have gcd(6,x) = 3, so x is divisible by 3 but not by 2.
Given LCM(18,24,x) = LCM(72,x) = 360 = 2^3·3^2·5, x must supply the factor 5 but must not introduce any prime power larger than those in 360. Since x is not divisible by 2, its form must be 3^a·5 with a = 1 or 2.
Thus x = 3·5 = 15 or x = 3^2·5 = 45.
Answer: 15 or 45.
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