Question

Just a simple yet interesting mathematical puzzle

Answer 1

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Answer 2

If you observe a number "3025" closely, you would find out that if this number was divided from middle you would get 2 numbers which when added and then squared would give back the original number.

(30+25)^2 = 3025

Find another such number with 4 digits all different.

Answer 3

@ganeshie8

Answer 4

i think im going blind

Answer 5

let that be abcd

\(10^4 \times a + 10^3 \times b + 10^2 \times c + 10 \times d\)

Answer 6

Well @mathslover : that number must be :

\[10^3 * a + 10^2 * b + 10*c + d\]

Answer 7

oh yeah, sorry!

Answer 8

@hartnn @Miracrown @iambatman @Kainui

Answer 9

(10a + b + 10c + d)^2 = (1000 a + 100 b + 10 c + d)

Answer 10

Good

Answer 11

\((10(a+c) + (b+d))^2 = 1000a + 100 b + 10 c + d\)

makes much more sense now.

Answer 12

Okay :)

Answer 13

100(a+c)^2 + (b+d)^2 + 20(a+c)(b+d) = 1000a + 100b + 10c + d

100a^2 + 100c^2 + 200ac + b^2 + d^2 +2bd + 20(a+c)(b+d) = 1000a + 100b + 10 c + d

(1000a-100a^2 ) + (100b - b^2) + (10c - 100c^2) + (d-d^2) = 200ac + 2bd + 20(a+c)(b+d)

100a(10 - a) + b(100 - b) + 10c(1 - 10c) + d(1 -d) = 200 ac + 2bd + 20(a+c)(b+d)

i will check whether i am doing right or not

we had 3025
a =3 ; b = 0 ; c = 2 ; d = 5

100(3)(7) + 0 + 20(1-20) + 5(-4) = 200(3)(2) + 0 + 20(3 + 2)(5)

2100 - 380 - 20 = 1200 + 500

1700 = 1700

that means, I'm doing right.

Will get back to you after dinner. :)

Answer 14

This would be allot faster if you applied the "Chinese method".

Answer 15

Got it.. :)

Answer 16

Okay I think I have given a lot of time. So, here is the solution:

The other number that answers all the requirements of the puzzle is 9,801. If we divide this in the middle into two numbers and add them together we get 99, which, multiplied by itself, produces 9,801. It is true that 2,025 may be treated in the same way, only this number is excluded by the condition which requires that no two figures should be alike.


The general solution:

(1) Call the number of figures in each half of the torn label n.

(2) Then, if we add 1 to each of the exponents of the prime factors (other than 3) of \(\large{10^n - 1}\) (1 being regarded as a factor with the constant exponent, 1), their product will be the number of solutions.

(3) Thus, for a label of six figures, n = 3.

The factors of \(\large{10^3 - 1}\) are \(\large{1^1 × 37^1}\) (not considering the \(\large{3^3}\)), and the product of \(\large{2 \times 2 = 4}\), the number of solutions.

This always includes the special cases 98 - 01, 00 - 01, 998 - 01, 000 - 001, etc.

The solutions are obtained as follows:—

Factorize \(\large{10^3 - 1}\) in all possible ways, always keeping the powers of 3 together, thus, \(\large{37 \times 27, 999 \times 1}\).

Then, solve the equation

\(\large{37x = 27y +1}\)

Here x = 19 and y = 26.

Therefore, \(\large{19 \times 37 = 703}\), the square of which gives one label, \(\large{494,209}\).

A complementary solution (through \(\large{27x = 37x + 1}\)) can at once be found by

\(\large{10^n - 703 = 297}\),

the square of which gives 088,209 for second label. (These non-significant zeroes to the left must be included, though they lead to peculiar cases like 00238 - 04641 = 4879²,
where 0238 - 4641 would not work.)

The special case \(\large{999 \times 1}\) we can write at once 998,001, according to the law shown above, by adding nines on one half and zeroes on the other, and its complementary will be 1 preceded by five zeroes, or 000001.

Thus we get the squares of 999 and 1.

These are the four solutions.

Answer 17

A more complex but short solution. :)

Answer 18

@hartnn @Miracrown @ganeshie8 @Kainui @mathmale

Answer 19

I will have to study your method. But, before that, I did something :

\(100a(10-a) + b(100-b) + 10c(1-10c) + d(1-d) = 20(ab+cd) \\
(ab+cd) = 5a(10-a) + b(100-b)/20 + c(1-10c)/2 + d(1-d)/20 = ab + cd \)

Answer 20

While here, in the RHS, ab and cd doesn't mean the numbers infact, those are the multiplication of two digits a and b.

Answer 21

Yeah but you have to remember that may result even in two digit numbers.

Answer 22

I will take some conditions into consideration now for a, b, c and d. As, not every number can be "d" .

Answer 23

I know, I'm making it a lot complex but I'm really interested in doing this problem with some new stuff.

Answer 24

naah... :(

Answer 25

Well, many are. I just be chance got the answer as I have solved a very similar question before.

Answer 26

Can we take some conditions here? Like I mentioned before, not every number can be d.

d can be 0,1,4,9,5,6

Answer 27

Oh, well, the thinker @Kainui is here. Let's see what he has got for us... :-)

Answer 28

Well see the problem is that you know the possible digits of a squared number so these digits must be resulted from squaring the sum of unit digits of each torn number.

Answer 29

Yeah.

Answer 30

I am fascinated by this and all questions that deal with digits. I don't have a very strong intuition in this area. I'm playing around with this right now though, it's interesting.

Answer 31

Thanks @Kainui.

Answer 32

I think I might have found an alternate way to solve this. One sec let me make sure.

Answer 33

Take your time :)

Answer 34

I've found some interesting things, but they've overall not ended well.

Answer 35

Well, that's ok. Thanks for trying though :)