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Question Number 22567 by mondodotto@gmail.com last updated on 20/Oct/17

Answered by Rasheed.Sindhi last updated on 20/Oct/17

$$840=2^3.3.5.7$$$$\therefore \mathrm{The}\mathrm{numbers}\left(\mathrm{of}\mathrm{which}840\mathrm{is}\mathrm{lcm}\right)$$$$\mathrm{may}\mathrm{have}\mathrm{factors}2,3,5,7\mathrm{all}\mathrm{or}\mathrm{some}$$$$\mathrm{of}\mathrm{them}.$$$$\mathrm{Suppose}\mathrm{the}\mathrm{third}\mathrm{number}\mathrm{is}\left(\mathrm{say}\mathrm{y}\right)$$$$2^{\mathrm{a}}.3^{\mathrm{b}}.5^{\mathrm{c}}.7^{\mathrm{d}}\mathrm{with}\mathrm{the}\mathrm{possibility}\mathrm{that}$$$$\mathrm{some}\mathrm{of}\mathrm{a},\mathrm{b},\mathrm{c},\mathrm{d}\mathrm{may}\mathrm{equal}\mathrm{to}0.$$$$24=2^3.3,42=2.3.7,\mathrm{y}=2^{\mathrm{a}}.3^{\mathrm{b}}.5^{\mathrm{c}}.7^{\mathrm{d}}$$$$\mathrm{HCF}$$$$=2^{\min (3,1,\mathrm{a})}.3^{\min (1,\mathrm{b})}.5^0.7^0=2^1.3^1$$$$\min (1,\mathrm{a})=1\Rightarrow \mathrm{a}⩾1..............\left(\mathrm{i}\right)$$$$\min (1,\mathrm{b})=1\Rightarrow \mathrm{b}⩾1.............\left(\mathrm{ii}\right)$$$$\mathrm{LCM}$$$$=2^{\max (3,1,\mathrm{a})}.3^{\max (1,\mathrm{b})}.5^{\max (0,\mathrm{c})}.7^{\max (0,1,\mathrm{d})}$$$$=2^3.3.5.7$$$$\max (3,1,\mathrm{a})=3\Rightarrow \mathrm{a}⩽3..............\left(\mathrm{iii}\right)$$$$\max (1,\mathrm{b})=1\Rightarrow \mathrm{b}⩽1..................\left(\mathrm{iv}\right)$$$$\max (0,\mathrm{c})=1\Rightarrow \mathrm{c}=1...................\left(\mathrm{v}\right)$$$$\max (0,1,\mathrm{d})=1\Rightarrow \mathrm{d}⩽1.............\left(\mathrm{vi}\right)$$$$\left(\mathrm{i}\right)\mathrm{\&}\left(\mathrm{iii}\right):\mathrm{a}⩾1\wedge \mathrm{a}⩽3\Rightarrow \mathrm{a}=1,2,3$$$$\left(\mathrm{ii}\right)\mathrm{\&}\left(\mathrm{iv}\right):\mathrm{b}⩾1\wedge \mathrm{b}⩽1\Rightarrow \mathrm{b}=1$$$$\left(\mathrm{v}\right):\mathrm{c}=1$$$$\left(\mathrm{vi}\right):\mathrm{d}⩽1^{}\Rightarrow \mathrm{d}=0,1$$$$\mathrm{Hence}\mathrm{the}\mathrm{third}\mathrm{number}\mathrm{is}$$$$2^{1,2,3}.3^1.5^1.7^{0,1}$$$$2^1.3.5.7^0=30$$$$2^1.3.5.7^1=210$$$$2^2.3.5.7^0=60$$$$2^2.3.5.7^1=420$$$$2^3.3.5.7^0=120$$$$2^3.3.5.7^1=840$$

Answered by Rasheed.Sindhi last updated on 22/Oct/17

$$\mathrm{A}\mathrm{simple}\mathrm{approach}$$$$\mathrm{Let}\mathrm{the}\mathrm{third}\mathrm{number}\mathrm{is}\mathrm{y}.$$$$\frac{24}{6}\times \frac{42}{6}\times \frac{\mathrm{y}}{6}\times 6=840$$$$\mathrm{y}=30$$$$\mathrm{The}\mathrm{lowest}\mathrm{answer}.\mathrm{Other}\mathrm{answers}$$$$\mathrm{are}\mathrm{such}\mathrm{multiples}\mathrm{of}\mathrm{this}\mathrm{which}$$$$\mathrm{assure}\mathrm{the}\mathrm{following}$$$$\left(\mathrm{i}\right)\mathrm{lcm}(24,42,\mathrm{y})=840$$$$\mathrm{and}$$$$\left(\mathrm{ii}\right)\gcd (24,42,\mathrm{y})=6$$

Answered by Rasheed.Sindhi last updated on 23/Oct/17

$$\mathrm{An}\mathrm{other}\mathrm{simple}\mathrm{approach}$$$$\mathrm{lcm}=840\mathrm{and}\gcd =6$$$$\mathrm{Let}\mathrm{the}\mathrm{third}\mathrm{is}\mathrm{t}$$$$\mathrm{lcm}(24,42,\mathrm{t})=840$$$$\mathrm{lcm}(2^3.3,2.3.7,2^{\mathrm{a}}.3^{\mathrm{b}}.5^{\mathrm{c}}.7^{\mathrm{d}})=2^3.3^{}.5.7$$$$\mathrm{a}⩽3,\mathrm{b}⩽1,\mathrm{c}=1,\mathrm{d}⩽1...............\left(\mathrm{i}\right)$$$$$$$$\gcd (2^3.3,2.3.7,2^{\mathrm{a}}.3^{\mathrm{b}}.5^{\mathrm{c}}.7^{\mathrm{d}})=2.3$$$$\mathrm{a}⩾1,\mathrm{b}⩾1,\mathrm{c}⩾0,\mathrm{d}⩾0...............\left(\mathrm{ii}\right)$$$$$$$$\left(\mathrm{i}\right)\mathrm{\&}\left(\mathrm{ii}\right):1⩽\mathrm{a}⩽3,\mathrm{b}=1,\mathrm{c}=1,\mathrm{d}⩽1$$$$\mathrm{a}=1,2,3;\mathrm{b}=1;\mathrm{c}=1;\mathrm{d}⩽1$$$$\mathrm{t}=2^{1,2,3}.3.5.7^{0,1}$$$$=2^1.3.5.7^0=30$$$$=2^1.3.5.7^1=210$$$$=2^2.3.5.7^0=60$$$$=2^2.3.5.7^1=420$$$$=2^3.3.5.7^0=120$$$$=2^3.3.5.7^1=840$$

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