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xholicwriter
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find a 4-digit phone number, given that it is a square. If we add 1 unit to each digit of the phone number, the new number we obtained is also a square.
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Char. Limit said:Oh, I see now. Starting with the equation n2 + 1111 = m2, we can subtract n2 from both sides and then factor to obtain 1111=(m+n)(m-n). Now, since m and n are integers, m+n and m-n must also be integers, and therefore factors of 1111. 1111 has only two factors: 11 and 101. Therefore, m+n and m-n must equal 101 and 11, respectively. (assuming that both m and n are positive, and that m>n, which is obvious from the original problem)
So now we have two equations: m+n=101 and m-n=11. Two distinct equations and two unknowns mean that there is exactly one solution. Let's add the two equations to get 2m=112. Therefore, m=56. Substitute m=56 into m-n=11 to get 56=n+11 and therefore n=45.
Finally, square n to get the original number, n2, which is 2025 and the 4-digit number that solves the problem.
xholicwriter said:Cool! if you like this type of problem, try the next one/ find a 5-digit number. If we take off the last 3 digits, the new number we obtained is the cube of the original number
micromass said:You want to find a two digit number n such that
[tex]0<n^3-1000n<1000[/tex]
The first inequality yields that [itex]n>\sqrt{1000}>31[/itex]. The second inequality requires us to find an n such that
[tex]n^3-1000n-1000<0[/tex]
The derivative of this is
[tex]3n^2-1000[/tex]
and thus the original function has a minimum in (approximatly) 18 and that the function is negative there. After 18, it rises. So the only possible n's are the ones following right after 31. It turns out that 32 was the only answer.
xholicwriter said:AWESOME! how can you figure out the solution in such a short period of time? It took me 30' to find out the answer :(
micromass said:Sorry
It's experience, I think...
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