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Tuesday, March 29, 2011
CAT Number Theory - Challenging question
Just one question this time, but a fairly challenging one though.
There is a 4-digit number 'abcd' that satisfies the following property. 'abcd' = ab ^2 + cd ^2. Find abcd.
Let the four digit number abcd = XY such that XY = X^2 + Y^2 => 100X + Y = X^2 + Y^2 => (X - 50)^2 + (Y - 1/2)^2 = 10001/4 => (X - 50)^2 + {(2Y - 1)^2}/4 = 10001/4 => 4(X - 50)^2 + (2Y - 1)^2 = 10001
Now sum of last two digits of two perfect squares is 01, it is only possible for the following two combinations; (00 + 01) or (76 + 25). It can be easily checked that first combination doesn't help and in the second one also there is only one favorable case i.e. 76^2 + 65^2 = 5776 + 4225 = 10001. So X = 12 and Y = 33 and the four digit number = abcd = XY = 1233 = 12^2 + 33^2.
When its 0, LHS will have two zeros at the end, so not possible
When its 2, LHS can never have 2 as unit digit, so not possible
When its 6, cd can have unit digit as 3 or 8
Hence, we can write as x(x - 1) = p(100 - p) = y, where x = cd, p = ab and y has a unit digit of 6 => x² - x - y = 0 Discriminant = 4y + 1 Unit digit of D will be 5, since it is a perfect square its ten's digit should be 2.
=> Tens digit of y should be 5 or 0, so last two digits of y should be 06 or 56.
Since p has unit digit 2 or 8, one of p and (100 - p) will be q2 and other will be (9 - q)8
Last two digits of (9 - q)8*q2 = 06 or 56 => Unit digit of 2(9 - q) + 8q will be 4 => Unit digit of 18 + 6q will be 4 => q can be 1 or 6
9 comments:
1233 = 12^2 + 33^2
Let the four digit number abcd = XY such that XY = X^2 + Y^2
=> 100X + Y = X^2 + Y^2
=> (X - 50)^2 + (Y - 1/2)^2 = 10001/4
=> (X - 50)^2 + {(2Y - 1)^2}/4 = 10001/4
=> 4(X - 50)^2 + (2Y - 1)^2 = 10001
Now sum of last two digits of two perfect squares is 01, it is only possible for the following two combinations; (00 + 01) or (76 + 25).
It can be easily checked that first combination doesn't help and in the second one also there is only one favorable case i.e. 76^2 + 65^2 = 5776 + 4225 = 10001. So X = 12 and Y = 33 and the four digit number = abcd = XY = 1233 = 12^2 + 33^2.
Kamal Lohia
abcd = ab² + cd²
=> ab(100 - ab) = cd(cd - 1)
Now, unit digit of RHS can be 0 or 2 or 6,
When its 0, LHS will have two zeros at the end, so not possible
When its 2, LHS can never have 2 as unit digit, so not possible
When its 6, cd can have unit digit as 3 or 8
Hence, we can write as
x(x - 1) = p(100 - p) = y, where x = cd, p = ab and y has a unit digit of 6
=> x² - x - y = 0
Discriminant = 4y + 1
Unit digit of D will be 5, since it is a perfect square its ten's digit should be 2.
=> Tens digit of y should be 5 or 0, so last two digits of y should be 06 or 56.
Since p has unit digit 2 or 8, one of p and (100 - p) will be q2 and other will be (9 - q)8
Last two digits of (9 - q)8*q2 = 06 or 56
=> Unit digit of 2(9 - q) + 8q will be 4
=> Unit digit of 18 + 6q will be 4
=> q can be 1 or 6
=> ab = 12(cd = 33) or 62(not possible)
So, only such number is 1233.
Rajesh,
I am not very sure but just wondering if there is a solution for this.Please post the solution if you have .
max value of ab ^2 + cd ^2 is1458 when abcd = 9999 implies
"a has to be 1considering abcd a four digit number"
so with the value of a as1 the maximum value of ab ^2 + cd ^2 =810 when abcd =1999
so according to me there is no solution for abcd =ab ^2 + cd ^2
Please correct me if i am wrong.
-Sindhu
121212....(300 times)/99
then would be the remainder?
10000=100^2+00^2
Wow excellent !!!
So which is correct answer whether 1233 or 10000.? Please reply me, I would like to know
its a 4 digit no....so 10000 is ruled out buddy. 1233...great technique kamal :)
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