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Question Number 116658 by zakirullah last updated on 05/Oct/20

Commented by Rasheed.Sindhi last updated on 08/Oct/20

$$\mathbb{A}\mathrm{n}\mathbb{O}\mathrm{ther}\mathbb{W}\mathrm{ay}\curvearrowleft \underset{\downarrow }{\uparrow }\curvearrowright$$$$12=2^2.3$$$$Thesmallestperfectsquare$$$$divisibleby12:2^2.3.3=36$$$$16=2^4$$$$Thesmallestperfectsquare$$$$divisibleby16:2^4=16$$$$20=2^2.5$$$$Thesmallestperfectsquare$$$$divisibleby20:2^2.5.5=100$$$$24=2^3.3$$$$Thesmallestperfectsquare$$$$divisibleby24:2^3.3.2.3=144$$$$Thesmallestperfectsquare$$$$numberwhichisdivisible$$$$by12,16,20\mathrm{\&}24:$$$$\mathrm{LCM}(36,16,100,144)=3600$$

Commented by zakirullah last updated on 08/Oct/20

$$\mathit{t}\mathit{h}\mathit{a}\mathit{n}\mathit{k}\mathit{s}\mathit{o}\mathit{m}\mathit{u}\mathit{c}\mathit{h}\mathit{s}\mathit{i}\mathit{r}$$

Answered by mr W last updated on 05/Oct/20

$$12=2^2\times 3$$$$16=2^4$$$$20=2^2\times 5$$$$24=2^3\times 3$$$$\Rightarrow 2^4\times 3^2\times 5^2=3600$$

Commented by zakirullah last updated on 05/Oct/20

$$thanksalot$$

Commented by zakirullah last updated on 05/Oct/20

$$sirhow{5}^{2}comes$$

Commented by JDamian last updated on 05/Oct/20

it is on purpose. Otherwise the number wouldn't be a perfect squared. In fact, all the primes of the number must be present as an even power.

Answered by $@y@m last updated on 05/Oct/20

$$12=2\times 2\times 3$$$$16=2\times 2\times 2\times 2$$$$20=2\times 2\times 5$$$$24=2\times 2\times 2\times 3$$$$Allexcept5havepair.$$$$Thereforerequiredno.is$$$$2\times 2\times 2\times 2\times 3\times 3\times 5\times 5=3600$$

Commented by zakirullah last updated on 05/Oct/20

$$siransis3600$$

Commented by zakirullah last updated on 05/Oct/20

$$thanks$$

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