Question

If a, b are natural numbers such that 2013 + a^2 = b^2, then the minimum possible value of ab is
(A) 671 (B) 668 (C) 658 (D) 645

Answer 1

http://en.wikipedia.org/wiki/Triangle_inequality Use the triangle inequality.

Answer 2

But how to find the minimum value?

Answer 3

Let the product of a and b be called P.
P = a * b ---- (1)
2013 + a^2 = b^2 ----- (2)
Find b from (2), put in (1).
You will have P as a function of just "a".

Find dP/da, equate it to 0, solve for "a". Make sure it gives a minimum.

Answer 4

^^^ not working for me. :(

Answer 5

yeahh it gives a = b i think... which wont fit the equation

Answer 6

The natural numbers part is throwing me off. Short of spamming 1 through sqrt(2013) until you find 2 natural numbers that fit, I'm drawing a blank.

Answer 7

First think I notice is ab will grow when a or b grow. So the first natural number that makes 2013 + a^2 a square number will give a minimum product. Through trial and error, I found 14 and 47 work

14 * 47 = 658

Answer 8

First *thing*

Answer 9

This a list of some possible values of (a,b)
{{14, 47}, {86, 97}, {334, 337}, {1006, 1007}}

Answer 10

The min is 47(14)=658

Answer 11

Any other way to solve this problem besides trial and error?

Answer 12

It is a Lagrange Multiplier problem
Minimize f(a,b) = a b
subject to
\[
a^2 + 2013 = b^2
\]

Answer 13

I tried Lagrange Multiplier. Didn't seem to work

Answer 14

The answer is given as 658.Could u tell how to solve it by Lagrange Multiplier .

Answer 15

I think Lagrange Multipliers does not seem to work with me.

Answer 16

We do not have to forget that a and b are integers and not any real numbers.
I wrote a mathematica script to generate my list

Answer 17

I ran it from a=1 to 100000

Answer 18

Here is the script
Q = {};
For[a = 1, a <= 100000, a++,
b = Sqrt[2013 + a^2];
If[ IntegerQ[ b] , Q = Append[Q, {a, b}]]]
Q

Answer 19

(b – -a) (b + a) = 2013 = 3 × 11 × 61
ab minimum when b -– a = 33
b + a = 61
a = 14

Answer 20

But why b-a is minimum when it is 33,why not 3,11 or 61.

Answer 21

The possibilities are:
(b-a) (b+a)
1 3x11x61
3 11 x 61
11 3 x 61
61 3 x 11 (not possible)
3x11 61
3x61 11 (not possible)
11x61 3 (not possible)
3x11x61 1 (not possible)

Answer 22

you have have only 3 possibilities.
66 x 61
3 x 617
183 x 11

try each pair and see which one gives positive integer. As I said earlier, the first lowest integer will give the minimum product

Answer 23

whoops i meant
33 x 61

Answer 24

hold on, I messed up multiplicatoin

Answer 25

it's
33 x 61
3 x 617
183 x 11

Answer 26

(33 + 61)/2 = 47
(3 + 617)/2 = 310
(183 + 11)/2 = 97

Answer 27

so the other solution is (47^2 - 2013)^.5 = 14

Answer 28

Here is another thought, define
\[
f(a )=a \sqrt{a^2+2013}
\]
This function is always increasing. At a=14 , f(14) = 658.
a=14 is the first value that makes f(a) an integer.
That does it

Answer 29

it still involves trial and errors but at least it's still better than blindly use the calculator and change each natural number

Answer 30

The reason the earlier calculus methods did not work for this problem is because this is a discrete function. It is not continuous and cannot be differentiated to find the minima.

Answer 31

This shows that f(14) is a minimum for integer values

Answer 32

The discrete part is subset of the continuous version

Answer 33

You only have to tests 14 numbers to get your result. That is not a lot even by hand.

Answer 34

Also, \(\large f(a )=a \sqrt{a^2+2013} \)

forces \(\large a^2+2013\) to be a perfect square greater than 44 : 45/46/47

Answer 35

we get lucky in two trials for sqrt(a^2+2013) = 45, 46, 47...
47 works !

Answer 36

Refer to the Mathematica attachement.