A number contains 8 digits. If you subtract the first 4 digits from the square of the final four, you obtain the original number.What is it ? A 7 dig example is 1201096. BREAK THE 8 NUMBER INTO 2 units. 1,000 a + 100b + 10c + d (1,000v + 100x + 10y + z) ^ 2. Would you need a computer to finish it ? Cheers.
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- Thread starter dave7
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A number contains 8 digits. If you subtract the first 4 digits from the square of the final four, you obtain the original number.What is it ? A 7 dig example is 1201096. BREAK THE 8 NUMBER INTO 2 units. 1,000 a + 100b + 10c + d (1,000v + 100x + 10y + z) ^ 2. Would you need a computer to finish it ? Cheers.
I do not understand that.
Original number assumed to be abcdvxyz
then the given condition is (I assume - we maintain the same order):
-(103 * a + 102 * b + 10 * c + d) + (103 * v + 102 * x + 10 * y + z)2 = 107 * a + 106 * b + 105 * c + 104d + 103 * v + 102 * x + 10 * y + z
I assume one restriction would be that all the number are positive integers.
In the absence of any other restriction, there are multiple answers to the problem. Yes, I would use programming to track that down.
One trivial answers is: 00000000 (if 02 is defined to be equal to 0)
Hello, dave7!
I have a set-up foir this problem . . . the rest is up to you.
Let \(\displaystyle N\) be the 8-digit number.
Let \(\displaystyle A\) be the 4-digit number formed by the first 4 digits;
. . let \(\displaystyle B\) be the 4-digit number formed by the last 4 digits.
Then \(\displaystyle N \:=\:10^4A + B\)
We are told that: .,\(\displaystyle B^2 - A \:=\:10^4A+B\)
And we have a choice of two equations to solve . . .
\(\displaystyle [1]\;A \:=\:\dfrac{B(B-1)}{10,\!001}\)
. . .where the product of two consecutive integers is a multiple of \(\displaystyle 73\!\cdot\!137.\)
\(\displaystyle [2]\;B^2 - B - 10,\!~001A \:=\:0 \quad\Rightarrow\quad B \:=\:\dfrac{1 \pm \sqrt{1+40,\!004A}}{2}\)
. . where \(\displaystyle 1 + 40,\!004A\) is a square.
Good luck!
I have a set-up foir this problem . . . the rest is up to you.
A number contains 8 digits.
If you subtract the first 4 digits from the square of the final four, you obtain the original number.
What is the number?
Let \(\displaystyle N\) be the 8-digit number.
Let \(\displaystyle A\) be the 4-digit number formed by the first 4 digits;
. . let \(\displaystyle B\) be the 4-digit number formed by the last 4 digits.
Then \(\displaystyle N \:=\:10^4A + B\)
We are told that: .,\(\displaystyle B^2 - A \:=\:10^4A+B\)
And we have a choice of two equations to solve . . .
\(\displaystyle [1]\;A \:=\:\dfrac{B(B-1)}{10,\!001}\)
. . .where the product of two consecutive integers is a multiple of \(\displaystyle 73\!\cdot\!137.\)
\(\displaystyle [2]\;B^2 - B - 10,\!~001A \:=\:0 \quad\Rightarrow\quad B \:=\:\dfrac{1 \pm \sqrt{1+40,\!004A}}{2}\)
. . where \(\displaystyle 1 + 40,\!004A\) is a square.
Good luck!