A three digit number

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A three digit number

Find a three digit number such that when raised to power 2, 3, 4, and all other integral powers, the products will end up with the original three digits. For example, if the three digits are 456, then all integral powers of 456 should end up with 456. (HINT: There are only two sets of three digit numbers having the specified property). Good luck. R.B.

Posted 137 day ago

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	Best Answer Here is my solution: Assume your three digit number is N, and its last two digits is S. Since power 2, 3, etc of N must end like the original three digits, it becomes obvious that N^2 – N must end with 3 zeros. Also, S^2 – S must end with two zeros. Now, we know that numbers ending with 5 or 6 when raised to any power will have their last digit 5 or 6. This is going to be our starting point. What about numbers ending with 25? Obviously, all numbers ending with 25, when raised to any power will have their last two digits 25 as well. So, one of our three digit numbers must end with 25! Trying 125, 225, … ,925, we see that only 625^2 – 625 = 390,000 ends with 3 zeros. Therefore 625 is one of the three digit numbers we were trying to find. Now, our second 3-digit number has a 6 as its last digits. Trying 16, 26, 36, ---, 96, we see that only 76^2 – 76 ends with two zeros. So our second 3-digit number will end with 76. Again trying 176, 276, 376, …, 976 we find that only 376^2 – 376 = 141,000. And our solution is complete. We found 625, and 376 as our numbers satisfying the specified conditions. Was very interesting problem indeed! R.B. Posted 135 day ago ( 0 ) ( 0 ) Report Abuse

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Answers (9)

To: matherapist:
Please read my solution carefully. I have covered everything that you want to see in the solution. Thank you. R.B.

Posted 135 day ago

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matherapist

Raj,

You need to give the rationale for the other integral powers. It's clear that if the last 3 digits are x (376 or 625) then any power will form the last 3 digits as the square.

Posted 135 day ago

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Here is my solution:
Assume your three digit number is N, and its last two digits is S. Since power 2, 3, etc of N must end like the original three digits, it becomes obvious that N^2 – N must end with 3 zeros. Also, S^2 – S must end with two zeros. Now, we know that numbers ending with 5 or 6 when raised to any power will have their last digit 5 or 6. This is going to be our starting point. What about numbers ending with 25? Obviously, all numbers ending with 25, when raised to any power will have their last two digits 25 as well. So, one of our three digit numbers must end with 25! Trying 125, 225, … ,925, we see that only 625^2 – 625 = 390,000 ends with 3 zeros. Therefore 625 is one of the three digit numbers we were trying to find.
Now, our second 3-digit number has a 6 as its last digits. Trying 16, 26, 36, ---, 96, we see that only 76^2 – 76 ends with two zeros. So our second 3-digit number will end with 76. Again trying 176, 276, 376, …, 976 we find that only 376^2 – 376 = 141,000. And our solution is complete. We found 625, and 376 as our numbers satisfying the specified conditions. Was very interesting problem indeed! R.B.

Posted 135 day ago

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matherapist

I didn't feel like expending the effort to do this analytically (actually I'm not sure I know how) so I used Brute force by modelling the problem on an excel spreadsheet. for x =1-999. I used x^2-x and x^3-x and checked for 3 zeroes at he end of each difference and then checked for x^4-x. Not very elegant but I did get . Took less than a minute to do the spreadsheet. and about 5 minutes to examine it.

376 and 625 as the numbers. Also 999 was a runner up (999^3-999) had 3 zeroes and 999^2-999 had 002.

Who thinks these things up?

Posted 136 day ago

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To: math_fanatic;

100 will not work, because when you raise it to power 2 or 3, , etc, the last 3 digits will not be 100, it will be 000. Do you see your mistake? Thank you. R.B.

Posted 136 day ago

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YummyCookies

i really dont care if u think i think that u r asking questions to get points! U ask too many questions and its really annoying, but i dont care at all and i just ignore! Btw, i really dont care if u r a retired math teacher or sumthing! Everyone is good at something, but if u try show off tht u r so good, thts not rite!

Posted 136 day ago

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math_fanatic

Are the digits all supposed to be different? Because 100 works.

Posted 136 day ago

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To: YummyCookies;
I am asking these questions to teach students and tutors some good mathematics. I don't need points. What am I going to do with more points? I am a retired math professor and I like to use my experience to teach others something useful. If you don't like my questions, just ignore them and do whatever you like to do.
Respectfully, Dr. Rajden Babayan

Posted 137 day ago

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YummyCookies

r u trying to ask questions to get more points?
if so, then thts not the rite thing t do!

Posted 137 day ago

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