Number Theory Aaggainnn...:(
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Is it possible to write a number using $1,2,3,....9$ digits once which is a perfect square and the last digit is $5$?
That is what I call"Real Number Theory Problem".
i.e:NO CONGRUENCE.
That is what I call"Real Number Theory Problem".
i.e:NO CONGRUENCE.
An amount of certain opposition is a great help to a man.Kites rise against,not with,the wind.
- Phlembac Adib Hasan
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Re: Number Theory Aaggainnn...:(
মানে বুঝলাম না। এক থেকে নয় পর্যন্ত কি সবগুলোই ব্যবহার করতে হবে নাকি মাঝে দুই একটা বাদ দেওয়া যাবে? বাদ দেওয়া গেলে তো একটা সহজ উত্তর আছে: $25$.
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Re: Number Theory Aaggainnn...:(
I think it means using the digits exactly once.
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Use $L^AT_EX$, It makes our work a lot easier!
Nur Muhammad Shafiullah | Mahi
Use $L^AT_EX$, It makes our work a lot easier!
Nur Muhammad Shafiullah | Mahi
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Re: Number Theory Aaggainnn...:(
Yes,It is...We have to use all the digits and exactly once.
An amount of certain opposition is a great help to a man.Kites rise against,not with,the wind.
- Phlembac Adib Hasan
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Re: Number Theory Aaggainnn...:(
এইটা কিছু হইল??? পুরোপুরি Bruit Force. কম্পিউটারের জন্য একেবারে আদর্শ। আমার তো ইচ্ছা করতেসে এখনই সি প্লাস প্লাসে একটা প্রোগ্রাম লিখে কম্পিউটারকে বসিয়ে দেই। RAM কম বলে অনেকটা সময় লাগবে। সেইজন্য খালি পারতেসি না।
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Re: Number Theory Aaggainnn...:(
Ans er ses 2 ta digit hobe 25
Re: Number Theory Aaggainnn...:(
এরকম কোন সংখ্যা নাই
- Fahim Shahriar
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Re: Number Theory Aaggainnn...:(
Suppose, there exists such number.
The number's last two digit will surely be $25$.
As its square root's last digit is $5$, then the third digit(from right) has 3 possible values; $0$,$2$ or $6$.
The number doesn't contain $0$ and $2$ has already been used as 2nd digit. So the third digit must be $6$.
Now, the number's last 3 digits are $625$ and its square root's last 2 digits are either $25$ and $75$.
The square root of the number contains 5 digits. We can write the number as $(10000a+1000b+100c+25)^2$ OR $(10000a+1000b+100c+75)^2$.
We will ignore $10000a+1000b$ as we are working only on 4th digit.
From $(100c+25)^2=10000c^2+5000c+625$ and $(100c+75)^2=10000c^2+15000c+5625$ , we can easily observe that the 4th digit will be either $0$ or $5$. But for same previous reasons it isn't possible.
সুতরাং এরকম কোন সংখ্যা নেই। Have I done any mistake?
The number's last two digit will surely be $25$.
As its square root's last digit is $5$, then the third digit(from right) has 3 possible values; $0$,$2$ or $6$.
The number doesn't contain $0$ and $2$ has already been used as 2nd digit. So the third digit must be $6$.
Now, the number's last 3 digits are $625$ and its square root's last 2 digits are either $25$ and $75$.
The square root of the number contains 5 digits. We can write the number as $(10000a+1000b+100c+25)^2$ OR $(10000a+1000b+100c+75)^2$.
We will ignore $10000a+1000b$ as we are working only on 4th digit.
From $(100c+25)^2=10000c^2+5000c+625$ and $(100c+75)^2=10000c^2+15000c+5625$ , we can easily observe that the 4th digit will be either $0$ or $5$. But for same previous reasons it isn't possible.
সুতরাং এরকম কোন সংখ্যা নেই। Have I done any mistake?
Name: Fahim Shahriar Shakkhor
Notre Dame College
Notre Dame College
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Re: Number Theory Aaggainnn...:(
Yes Fahim bhaiya is absolutely correct.
Yesterday is past, tomorrow is a mystery but today is a gift.
Re: Number Theory Aaggainnn...:(
This very nice problem is an example of why human brain is more powerful than a computer (cf. fahim's solution above).Phlembac Adib Hasan wrote:এইটা কিছু হইল??? পুরোপুরি Bruit Force. কম্পিউটারের জন্য একেবারে আদর্শ। আমার তো ইচ্ছা করতেসে এখনই সি প্লাস প্লাসে একটা প্রোগ্রাম লিখে কম্পিউটারকে বসিয়ে দেই। RAM কম বলে অনেকটা সময় লাগবে। সেইজন্য খালি পারতেসি না।
"Everything should be made as simple as possible, but not simpler." - Albert Einstein