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Question Number 206309 by BaliramKumar last updated on 11/Apr/24

Find the ways to express 11025 as product of two factors. (a) 13 (b) 14 (c) 26 (d) 27

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{ways}\:\mathrm{to}\:\mathrm{express}\:\mathrm{11025}\: \\ $$$$\mathrm{as}\:\mathrm{product}\:\mathrm{of}\:\mathrm{two}\:\mathrm{factors}. \\ $$$$\left(\mathrm{a}\right)\:\mathrm{13}\:\:\:\:\:\:\:\:\:\left(\mathrm{b}\right)\:\mathrm{14}\:\:\:\:\:\:\:\:\left(\mathrm{c}\right)\:\mathrm{26}\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{d}\right)\:\mathrm{27} \\ $$

Answered by TheHoneyCat last updated on 11/Apr/24

11025 =5×2205 =5^2 ×441 =3×5^2 ×147 =3^2 ×5^2 ×49 =3^2 ×5^2 ×7^2 From there, it is easy to count how many tuples (x,y) are such that xy=11025 This is 3×3×3=27 Let us now count how many of them are such that (x,y)=(y,x) x=3^a ×5^b ×7^c y=3^(2−a) ×5^(2−b) ×7^(2−b) ⇒a=b=c=1 so this is only one possibility. So there are 26 unsymetric tuples and 1 symetric leading to: 13+1 pairs {x,y} such that xy=11025 So overall: #{{x,y}∣xy=11025} = 14 _□

$$\mathrm{11025} \\ $$$$=\mathrm{5}×\mathrm{2205} \\ $$$$=\mathrm{5}^{\mathrm{2}} ×\mathrm{441} \\ $$$$=\mathrm{3}×\mathrm{5}^{\mathrm{2}} ×\mathrm{147} \\ $$$$=\mathrm{3}^{\mathrm{2}} ×\mathrm{5}^{\mathrm{2}} ×\mathrm{49} \\ $$$$=\mathrm{3}^{\mathrm{2}} ×\mathrm{5}^{\mathrm{2}} ×\mathrm{7}^{\mathrm{2}} \\ $$$$\mathrm{From}\:\mathrm{there},\:\mathrm{it}\:\mathrm{is}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{count}\:\mathrm{how}\:\mathrm{many} \\ $$$$\mathrm{tuples}\:\left({x},{y}\right)\:\mathrm{are}\:\mathrm{such}\:\mathrm{that}\:{xy}=\mathrm{11025} \\ $$$$\mathrm{This}\:\mathrm{is}\:\mathrm{3}×\mathrm{3}×\mathrm{3}=\mathrm{27} \\ $$$$\mathrm{Let}\:\mathrm{us}\:\mathrm{now}\:\mathrm{count}\:\mathrm{how}\:\mathrm{many}\:\mathrm{of}\:\mathrm{them} \\ $$$$\mathrm{are}\:\mathrm{such}\:\mathrm{that}\:\left({x},{y}\right)=\left({y},{x}\right) \\ $$$${x}=\mathrm{3}^{{a}} ×\mathrm{5}^{{b}} ×\mathrm{7}^{{c}} \\ $$$${y}=\mathrm{3}^{\mathrm{2}−{a}} ×\mathrm{5}^{\mathrm{2}−{b}} ×\mathrm{7}^{\mathrm{2}−{b}} \\ $$$$\Rightarrow{a}={b}={c}=\mathrm{1}\:\mathrm{so}\:\mathrm{this}\:\mathrm{is}\:\mathrm{only}\:\mathrm{one}\:\mathrm{possibility}. \\ $$$$\mathrm{So}\:\mathrm{there}\:\mathrm{are}\:\mathrm{26}\:\mathrm{unsymetric}\:\mathrm{tuples} \\ $$$$\mathrm{and}\:\mathrm{1}\:\mathrm{symetric} \\ $$$$\mathrm{leading}\:\mathrm{to}:\:\mathrm{13}+\mathrm{1}\:\mathrm{pairs}\:\left\{{x},{y}\right\}\:\mathrm{such}\:\mathrm{that} \\ $$$$\:{xy}=\mathrm{11025} \\ $$$$ \\ $$$$\mathrm{So}\:\mathrm{overall}: \\ $$$$#\left\{\left\{{x},{y}\right\}\mid{xy}=\mathrm{11025}\right\}\:=\:\mathrm{14}\:\:_{\Box} \\ $$

Commented by TheHoneyCat last updated on 11/Apr/24

By the way, for those who want to check, here is the full list in the form c,b,a ∣ x×y ∣ n with x≤y xy=11025 x=3^a 5^b 7^c n such that ∀x ∃! n

$$\mathrm{By}\:\mathrm{the}\:\mathrm{way},\:\mathrm{for}\:\mathrm{those}\:\mathrm{who}\:\mathrm{want}\:\mathrm{to}\:\mathrm{check}, \\ $$$$\mathrm{here}\:\mathrm{is}\:\mathrm{the}\:\mathrm{full}\:\mathrm{list}\:\mathrm{in}\:\mathrm{the}\:\mathrm{form} \\ $$$${c},{b},{a}\:\:\:\mid\:\:\:{x}×{y}\:\:\:\:\mid\:\:{n} \\ $$$$\mathrm{with} \\ $$$${x}\leqslant{y} \\ $$$${xy}=\mathrm{11025} \\ $$$${x}=\mathrm{3}^{{a}} \mathrm{5}^{{b}} \mathrm{7}^{{c}} \\ $$$${n}\:\mathrm{such}\:\mathrm{that}\:\forall{x}\:\exists!\:{n} \\ $$$$ \\ $$

Commented by TheHoneyCat last updated on 11/Apr/24

000 1×11025 1 001 3×3675 2 002 9×1125 5 010 5×2205 3 011 15×735 6 012 45×245 10 020 25×441 8 021 75×147 13 022 225 100 7×1575 4 101 21×525 7 102 63×175 12 110 35×245 9 111 105×105 14 ←Max Value 112 315 120 175 121 525 122 1575 200 49×255 11 201 147 202 441 210 245 211 735 212 2205 220 11225 221 3675 222 1125

$$\mathrm{000}\:\:\:\mathrm{1}×\mathrm{11025}\:\:\:\mathrm{1} \\ $$$$\mathrm{001}\:\:\:\mathrm{3}×\mathrm{3675}\:\:\:\:\:\:\mathrm{2} \\ $$$$\mathrm{002}\:\:\:\mathrm{9}×\mathrm{1125}\:\:\:\:\:\:\mathrm{5} \\ $$$$\mathrm{010}\:\:\:\mathrm{5}×\mathrm{2205}\:\:\:\:\:\:\mathrm{3} \\ $$$$\mathrm{011}\:\:\:\mathrm{15}×\mathrm{735}\:\:\:\:\:\:\mathrm{6} \\ $$$$\mathrm{012}\:\:\:\mathrm{45}×\mathrm{245}\:\:\:\:\:\:\mathrm{10} \\ $$$$\mathrm{020}\:\:\:\mathrm{25}×\mathrm{441}\:\:\:\:\:\:\mathrm{8} \\ $$$$\mathrm{021}\:\:\:\mathrm{75}×\mathrm{147}\:\:\:\:\:\mathrm{13} \\ $$$$\mathrm{022}\:\:\:\mathrm{225} \\ $$$$\mathrm{100}\:\:\:\mathrm{7}×\mathrm{1575}\:\:\:\:\:\mathrm{4} \\ $$$$\mathrm{101}\:\:\:\mathrm{21}×\mathrm{525}\:\:\:\:\:\mathrm{7} \\ $$$$\mathrm{102}\:\:\:\mathrm{63}×\mathrm{175}\:\:\:\:\:\mathrm{12} \\ $$$$\mathrm{110}\:\:\:\mathrm{35}×\mathrm{245}\:\:\:\:\:\:\mathrm{9} \\ $$$$\mathrm{111}\:\:\:\mathrm{105}×\mathrm{105}\:\:\:\mathrm{14}\:\:\:\:\:\:\leftarrow\mathrm{Max}\:\mathrm{Value} \\ $$$$\mathrm{112}\:\:\:\mathrm{315} \\ $$$$\mathrm{120}\:\:\:\mathrm{175} \\ $$$$\mathrm{121}\:\:\:\mathrm{525} \\ $$$$\mathrm{122}\:\:\:\mathrm{1575} \\ $$$$\mathrm{200}\:\:\:\mathrm{49}×\mathrm{255}\:\:\:\:\:\:\mathrm{11} \\ $$$$\mathrm{201}\:\:\:\mathrm{147} \\ $$$$\mathrm{202}\:\:\:\:\mathrm{441} \\ $$$$\mathrm{210}\:\:\:\:\mathrm{245} \\ $$$$\mathrm{211}\:\:\:\:\mathrm{735} \\ $$$$\mathrm{212}\:\:\:\:\mathrm{2205} \\ $$$$\mathrm{220}\:\:\:\:\mathrm{11225} \\ $$$$\mathrm{221}\:\:\:\:\mathrm{3675} \\ $$$$\mathrm{222}\:\:\:\mathrm{1125} \\ $$

Commented by BaliramKumar last updated on 12/Apr/24

Thanks

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Answered by mr W last updated on 11/Apr/24

11025=3^2 5^2 7^2 ⇒number of factors: 3×3×3=27 1+((27−1)/2)=14 ⇒14 ways

$$\mathrm{11025}=\mathrm{3}^{\mathrm{2}} \mathrm{5}^{\mathrm{2}} \mathrm{7}^{\mathrm{2}} \\ $$$$\Rightarrow{number}\:{of}\:{factors}:\:\mathrm{3}×\mathrm{3}×\mathrm{3}=\mathrm{27} \\ $$$$\mathrm{1}+\frac{\mathrm{27}−\mathrm{1}}{\mathrm{2}}=\mathrm{14} \\ $$$$\Rightarrow\mathrm{14}\:{ways} \\ $$

Commented by BaliramKumar last updated on 12/Apr/24

Thanks

$$\mathrm{Thanks} \\ $$

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