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Difficulty: HardHCF and LCM

Three different cryptographic algorithms in a secure server refresh their encryption keys at regular intervals. Algorithm X refreshes every 83\frac{8}{3} milliseconds, Algorithm Y every 109\frac{10}{9} milliseconds, and Algorithm Z every 145\frac{14}{5} milliseconds. They all start a refresh cycle simultaneously. Let MM be the minimum number of milliseconds until they all start a refresh cycle together again. Let HH be the Highest Common Factor (HCF) of their respective refresh intervals in milliseconds. What is the value of M÷HM \div H?

  1. A
    16300\frac{1}{6300}
  2. B
    140140
  3. C
    7840078400
  4. 63006300Answer

Answer

The value of M÷HM \div H is 63006300.
The value MM requires finding the LCM of the fractions, which represents the synchronization time. LCM(83,109,145)=LCM(8,10,14)HCF(3,9,5)=2801=280\text{LCM}(\frac{8}{3}, \frac{10}{9}, \frac{14}{5}) = \frac{\text{LCM}(8, 10, 14)}{\text{HCF}(3, 9, 5)} = \frac{280}{1} = 280. The value HH requires finding the HCF of the fractions. HCF(83,109,145)=HCF(8,10,14)LCM(3,9,5)=245\text{HCF}(\frac{8}{3}, \frac{10}{9}, \frac{14}{5}) = \frac{\text{HCF}(8, 10, 14)}{\text{LCM}(3, 9, 5)} = \frac{2}{45}. Finally, dividing MM by HH gives 280÷245=280×452=140×45=6300280 \div \frac{2}{45} = 280 \times \frac{45}{2} = 140 \times 45 = 6300.

Step-by-Step Solution

1
Determine the synchronization time MM by finding the LCM of the three fractional intervals.
M=LCM(83,109,145)M = \text{LCM}(\frac{8}{3}, \frac{10}{9}, \frac{14}{5})
Simultaneous repeating events synchronize at the Lowest Common Multiple of their individual cycle times.
2
Calculate the LCM using the fraction rule: LCM of numeratorsHCF of denominators\frac{\text{LCM of numerators}}{\text{HCF of denominators}}.
Numerators (8,10,14)(8, 10, 14) have an LCM of 280280. Denominators (3,9,5)(3, 9, 5) have an HCF of 11. Thus, M=2801=280M = \frac{280}{1} = 280.
This is the mathematical formula required for finding the LCM of rational numbers.
3
Determine the Highest Common Factor HH of the three fractional intervals.
H=HCF(83,109,145)H = \text{HCF}(\frac{8}{3}, \frac{10}{9}, \frac{14}{5})
The question explicitly requests the HCF of the three given refresh times.
4
Calculate the HCF using the fraction rule: HCF of numeratorsLCM of denominators\frac{\text{HCF of numerators}}{\text{LCM of denominators}}.
Numerators (8,10,14)(8, 10, 14) have an HCF of 22. Denominators (3,9,5)(3, 9, 5) have an LCM of 4545. Thus, H=245H = \frac{2}{45}.
This is the mathematical formula required for finding the HCF of rational numbers.
5
Calculate the final required ratio M÷HM \div H.
280÷245=280×452=140×45=6300280 \div \frac{2}{45} = 280 \times \frac{45}{2} = 140 \times 45 = 6300.
This satisfies the final computational step requested by the problem stem.

Key Concept

Calculating the Lowest Common Multiple (LCM) and Highest Common Factor (HCF) for fractions using their specific formulas.
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Topics

HCF and LCM

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