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Question Number 14757 by RasheedSoomro last updated on 04/Jun/17

Solve: ((1000!)/(5×10×15×...1000))≡x(mod 10)

$$\mathrm{Solve}: \\ $$$$\frac{\mathrm{1000}!}{\mathrm{5}×\mathrm{10}×\mathrm{15}×...\mathrm{1000}}\equiv\mathrm{x}\left(\mathrm{mod}\:\mathrm{10}\right) \\ $$

Answered by RasheedSoomro last updated on 05/Jun/17

x≡((1000!)/(5×10×15×...1000))(mod 10 x≡(1.2.3.4)(6.7.8.9)...(996.997.998.999)(mod 10) In brackets numbers are of 5k+1,5k+2,5k+3 and 5k+4 types. The following can be proved (5k+1)(5k+2)(5k+3)(5k+4)≡4(mod 10) (See answer of mrW1 to Q#14614) Hence 1.2.3.4≡4(mod 10) 6.7.8.9≡4(mod 10) ⋮ 996.997.998.999≡4(mod 10) Total congruences=200 (2 per each ten) Multiplying all the 200 congruences (1.2.3.4)(6.7.8.9)...(996.997.998.999)≡4^(200) ≡6(mod 10) (4^1 ≡4,4^2 ≡6,4^3 ≡4, ...,4^(2k+1) ≡4,4^(2k) ≡6(mod 10) ) So ((1000!)/(5×10×15×...1000))≡6 (mod 10)

$$\mathrm{x}\equiv\frac{\mathrm{1000}!}{\mathrm{5}×\mathrm{10}×\mathrm{15}×...\mathrm{1000}}\left(\mathrm{mod}\:\mathrm{10}\right. \\ $$$$\mathrm{x}\equiv\left(\mathrm{1}.\mathrm{2}.\mathrm{3}.\mathrm{4}\right)\left(\mathrm{6}.\mathrm{7}.\mathrm{8}.\mathrm{9}\right)...\left(\mathrm{996}.\mathrm{997}.\mathrm{998}.\mathrm{999}\right)\left(\mathrm{mod}\:\mathrm{10}\right) \\ $$$$\mathrm{In}\:\mathrm{brackets}\:\mathrm{numbers}\:\mathrm{are}\:\mathrm{of}\:\mathrm{5k}+\mathrm{1},\mathrm{5k}+\mathrm{2},\mathrm{5k}+\mathrm{3}\:\mathrm{and} \\ $$$$\mathrm{5k}+\mathrm{4}\:\mathrm{types}.\:\mathrm{The}\:\mathrm{following}\:\mathrm{can}\:\mathrm{be}\:\mathrm{proved} \\ $$$$\left(\mathrm{5k}+\mathrm{1}\right)\left(\mathrm{5k}+\mathrm{2}\right)\left(\mathrm{5k}+\mathrm{3}\right)\left(\mathrm{5k}+\mathrm{4}\right)\equiv\mathrm{4}\left(\mathrm{mod}\:\mathrm{10}\right) \\ $$$$\:\:\:\:\:\:\:\:\left(\mathrm{See}\:\mathrm{answer}\:\mathrm{of}\:\mathrm{mrW1}\:\mathrm{to}\:\mathrm{Q}#\mathrm{14614}\right) \\ $$$$\mathrm{Hence}\: \\ $$$$\:\:\:\mathrm{1}.\mathrm{2}.\mathrm{3}.\mathrm{4}\equiv\mathrm{4}\left(\mathrm{mod}\:\mathrm{10}\right) \\ $$$$\:\:\:\mathrm{6}.\mathrm{7}.\mathrm{8}.\mathrm{9}\equiv\mathrm{4}\left(\mathrm{mod}\:\mathrm{10}\right) \\ $$$$\:\:\:\:\vdots \\ $$$$\:\:\mathrm{996}.\mathrm{997}.\mathrm{998}.\mathrm{999}\equiv\mathrm{4}\left(\mathrm{mod}\:\mathrm{10}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{Total}\:\mathrm{congruences}=\mathrm{200}\:\left(\mathrm{2}\:\mathrm{per}\:\mathrm{each}\:\mathrm{ten}\right) \\ $$$$\mathrm{Multiplying}\:\mathrm{all}\:\mathrm{the}\:\mathrm{200}\:\mathrm{congruences} \\ $$$$\left(\mathrm{1}.\mathrm{2}.\mathrm{3}.\mathrm{4}\right)\left(\mathrm{6}.\mathrm{7}.\mathrm{8}.\mathrm{9}\right)...\left(\mathrm{996}.\mathrm{997}.\mathrm{998}.\mathrm{999}\right)\equiv\mathrm{4}^{\mathrm{200}} \equiv\mathrm{6}\left(\mathrm{mod}\:\mathrm{10}\right) \\ $$$$\left(\mathrm{4}^{\mathrm{1}} \equiv\mathrm{4},\mathrm{4}^{\mathrm{2}} \equiv\mathrm{6},\mathrm{4}^{\mathrm{3}} \equiv\mathrm{4},\:...,\mathrm{4}^{\mathrm{2k}+\mathrm{1}} \equiv\mathrm{4},\mathrm{4}^{\mathrm{2k}} \equiv\mathrm{6}\left(\mathrm{mod}\:\mathrm{10}\right)\:\:\right) \\ $$$$\mathrm{So}\:\frac{\mathrm{1000}!}{\mathrm{5}×\mathrm{10}×\mathrm{15}×...\mathrm{1000}}\equiv\mathrm{6}\:\left(\mathrm{mod}\:\mathrm{10}\right) \\ $$$$ \\ $$

Commented by RasheedSoomro last updated on 05/Jun/17

Now, I think, I′m able to answer the question of prakash jain. (To determine non-zero digit...) Q#13724

$$\mathrm{Now},\:\mathrm{I}\:\mathrm{think},\:\mathrm{I}'\mathrm{m}\:\mathrm{able}\:\mathrm{to}\:\mathrm{answer} \\ $$$$\mathrm{the}\:\mathrm{question}\:\mathrm{of}\:\boldsymbol{\mathrm{prakash}}\:\boldsymbol{\mathrm{jain}}. \\ $$$$\left(\mathrm{To}\:\mathrm{determine}\:\mathrm{non}-\mathrm{zero}\:\mathrm{digit}...\right) \\ $$$$\mathrm{Q}#\mathrm{13724} \\ $$

Commented by RasheedSoomro last updated on 05/Jun/17

To Sir prakash jain & mrW1 & All who are interested in Number Theory. Pl see my answer to your Q#13724 I have tried to solve. The solution is based on the above answer.

$$\mathrm{To}\:\boldsymbol{\mathrm{S}}\mathrm{ir}\:\mathrm{prakash}\:\mathrm{jain}\:\&\:\mathrm{mrW1} \\ $$$$\&\:\mathrm{All}\:\mathrm{who}\:\mathrm{are}\:\mathrm{interested}\:\mathrm{in}\:\boldsymbol{\mathrm{Number}}\:\boldsymbol{\mathrm{Theory}}. \\ $$$$\mathrm{Pl}\:\mathrm{see}\:\mathrm{my}\:\mathrm{answer}\:\mathrm{to}\:\mathrm{your}\:\mathrm{Q}#\mathrm{13724} \\ $$$$\mathrm{I}\:\mathrm{have}\:\mathrm{tried}\:\mathrm{to}\:\mathrm{solve}.\:\mathrm{The}\:\mathrm{solution}\:\mathrm{is} \\ $$$$\mathrm{based}\:\mathrm{on}\:\mathrm{the}\:\mathrm{above}\:\mathrm{answer}. \\ $$

Commented by mrW1 last updated on 05/Jun/17

nice work! ((1000!)/(5×10×15×...1000))≡6 (mod 10) I can confirm this through a different way: in 1000! there are 100 numbers which end with 1: 1,11,21,...101,111,...991 the product of them ends with 1 similarly there are 100 numbers ending with 2: 2,12,22,...,102,112,...,992 the product of them ends with 6 there are 100 numbers ending with 3, the product of them ends with 1 there are 100 numbers ending with 4, the product of them ends with 6 there are 100 numbers ending with 6, the product of them ends with 6 there are 100 numbers ending with 7, the product of them ends with 1 there are 100 numbers ending with 8, the product of them ends with 6 there are 100 numbers ending with 9, the product of them ends with 1 these values end with 1 or 6, so their product will also end with 6. that means if we exclude the numbers 10,20,...,100,110,...,990,1000 and 5,15,25,...,105,115,...,995, the product of the remaining numbers in 1000! will end with the digit 6.

$${nice}\:{work}! \\ $$$$ \\ $$$$\frac{\mathrm{1000}!}{\mathrm{5}×\mathrm{10}×\mathrm{15}×...\mathrm{1000}}\equiv\mathrm{6}\:\left(\mathrm{mod}\:\mathrm{10}\right) \\ $$$$ \\ $$$${I}\:{can}\:{confirm}\:{this}\:{through}\:{a}\:{different} \\ $$$${way}: \\ $$$$ \\ $$$${in}\:\mathrm{1000}!\:{there}\:{are}\:\mathrm{100}\:{numbers}\:{which} \\ $$$${end}\:{with}\:\mathrm{1}:\:\mathrm{1},\mathrm{11},\mathrm{21},...\mathrm{101},\mathrm{111},...\mathrm{991} \\ $$$${the}\:{product}\:{of}\:{them}\:{ends}\:{with}\:\mathrm{1} \\ $$$$ \\ $$$${similarly}\:{there}\:{are}\:\mathrm{100}\:{numbers}\:{ending} \\ $$$${with}\:\mathrm{2}:\:\mathrm{2},\mathrm{12},\mathrm{22},...,\mathrm{102},\mathrm{112},...,\mathrm{992} \\ $$$${the}\:{product}\:{of}\:{them}\:{ends}\:{with}\:\mathrm{6} \\ $$$$ \\ $$$${there}\:{are}\:\mathrm{100}\:{numbers}\:{ending}\:{with}\:\mathrm{3}, \\ $$$${the}\:{product}\:{of}\:{them}\:{ends}\:{with}\:\mathrm{1} \\ $$$$ \\ $$$${there}\:{are}\:\mathrm{100}\:{numbers}\:{ending}\:{with}\:\mathrm{4}, \\ $$$${the}\:{product}\:{of}\:{them}\:{ends}\:{with}\:\mathrm{6} \\ $$$$ \\ $$$${there}\:{are}\:\mathrm{100}\:{numbers}\:{ending}\:{with}\:\mathrm{6}, \\ $$$${the}\:{product}\:{of}\:{them}\:{ends}\:{with}\:\mathrm{6} \\ $$$$ \\ $$$${there}\:{are}\:\mathrm{100}\:{numbers}\:{ending}\:{with}\:\mathrm{7}, \\ $$$${the}\:{product}\:{of}\:{them}\:{ends}\:{with}\:\mathrm{1} \\ $$$$ \\ $$$${there}\:{are}\:\mathrm{100}\:{numbers}\:{ending}\:{with}\:\mathrm{8}, \\ $$$${the}\:{product}\:{of}\:{them}\:{ends}\:{with}\:\mathrm{6} \\ $$$$ \\ $$$${there}\:{are}\:\mathrm{100}\:{numbers}\:{ending}\:{with}\:\mathrm{9}, \\ $$$${the}\:{product}\:{of}\:{them}\:{ends}\:{with}\:\mathrm{1} \\ $$$$ \\ $$$${these}\:{values}\:{end}\:{with}\:\mathrm{1}\:{or}\:\mathrm{6},\:{so}\:{their} \\ $$$${product}\:{will}\:{also}\:{end}\:{with}\:\mathrm{6}. \\ $$$$ \\ $$$${that}\:{means}\:{if}\:{we}\:{exclude}\:{the}\:{numbers} \\ $$$$\mathrm{10},\mathrm{20},...,\mathrm{100},\mathrm{110},...,\mathrm{990},\mathrm{1000}\:{and} \\ $$$$\mathrm{5},\mathrm{15},\mathrm{25},...,\mathrm{105},\mathrm{115},...,\mathrm{995},\:{the}\:{product} \\ $$$${of}\:{the}\:{remaining}\:{numbers}\:{in}\:\mathrm{1000}!\: \\ $$$${will}\:{end}\:{with}\:{the}\:{digit}\:\mathrm{6}. \\ $$

Commented by RasheedSoomro last updated on 05/Jun/17

TH𝛂nksALOT Sir ! Nice&easy method!

$$\boldsymbol{\mathcal{TH}\alpha{nksA}\mathcal{LOT}}\:\:\boldsymbol{\mathcal{S}{ir}}\:! \\ $$$$\boldsymbol{\mathcal{N}{ice\&easy}}\:\:\boldsymbol{{method}}! \\ $$

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