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.

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9 comments

  1. 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

  2. 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.

  3. 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

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