Question 833791: If an integer ends on 0 or 5, then it is divisible by 5. If it ends on 0, 2, 4, 6 or 8, then it is divisible by 2. It is divisible by 3 if the sum of its digits is divisible by three.
Using these simple rules, determine the prime-factorings of 2295 and 2040. Then use these results to compute the prime-factorings of the following three numbers: 2295 * 2040, gcd(2295,2040) and lcm(2295,2040). Finally, demonstrate the relationship between these three numbers.
Answer by Edwin McCravy(20065)   (Show Source):
You can put this solution on YOUR website! If an integer ends on 0 or 5, then it is divisible by 5. If it ends on 0, 2, 4, 6 or 8, then it is divisible by 2. It is divisible by 3 if the sum of its digits is divisible by three.
Using these simple rules, determine the prime-factorings of 2295 and 2040. Then use these results to compute the prime-factorings of the following three numbers: 2295 * 2040, gcd(2295,2040) and lcm(2295,2040). Finally, demonstrate the relationship between these three numbers.
Prime factorization, "PF", of 2295 = 3×3×3×5×17 = 33×5×17
Prime factorization, "PF", of 2040 = 2×2×2×3×17 = 23×3×17
Line them up like this, putting in all factors of both numbers,
When there is a factor of only one of them, include that factor in
the other PF to the 0 power. Also include any 1 exponents:
PF(2295) = 20×33×51×171
PF(2040) = 23×31×51×171
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To get the prime factorization of their product, ADD the two exponents of
each factor and write the result below the line.
PF(2295) = 20×33×51×171
PF(2040) = 23×31×51×171
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PF(2295×2040) = 23×34×52×172 = 4681800
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To get their greatest common divisor, below the line, bring down each
factor with the MINIMUM exponent:
PF(2295) = 20×33×51×171
PF(2040) = 23×31×51×171
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gcd(2295×2040) = 20×31×51×171 = 255
###############################################################
To get their least common multiple, below the line bring down each
factor with the MAXIMUM exponent:
PF(2295) = 20×33×51×171
PF(2040) = 23×31×51×171
----------------------------
lcm(2295,2040) = 23×33×51×171 = 18360
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The relationship demonstrated here is that
A×B = gcd(A,B)×lcm(A,B)
2295×2040 = 4681800 = gcd(2295,2040)×lcm(2295,2040) = 255×18360 = 4681800.
Edwin
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