wei lun wrote:thks for your reply but need to use equation to solve this problem.

Hi wel lun,

Here you go:

Let the first digit be x,

Since the sum of the digits is 6, the second digit would be 6-x,

Therefore, the products of the digits would be = x(6-x)

= 6x-x^2

If the first digit is x, and it is a 2 digit number, x must be in the tens place, hence the NUMERICAL value of x would be = 10 x (x)

= 10x

Since the second digit is in the ones place, the NUMERICAL value of 6-x would be

= 1 x (6-x)

= 6-x

Hence, the NUMERICAL value of the 2 digit number would be = 10x + (6-x)

= 10x + 6 - x

= 9x +6

Since the products of the digits IS EQUAL to the 1/3 of the original number, we can construct the equation as follows,

6x-x^2 = (1/3) x (9x+6)

Rearranging the equation, we will have,

6x-x^2 = 3x+2

6x = x^2+3x+2

0 = x^2+3x-6x+2

0 = x^2-3x+2

x^2-3x+2 = 0

Factorising using the Cross technique you learnt in secondary 2,

You will get,

(x-2)(x-1)=0

x=2 or x=1 (first digit)

6-x=4 or x=5 (second digit)

Hence, the original number may be 24 or 15.

SOLVED! =D

You may call me at 92220737 if you require any further assistance in these maths questions, I will be glad to help! =)