~~{{{{一條英文數學題}}}~~[[[[緊急]]]]]____||||||||10點ar!!!|||||||

Only Number which is divisible by 10 can satisfy the condition..
so first digit is 0 (that eans the number is y0)

6y-( 10 y + 0)/2 = 4

6y is 6 times the ten digit
( 10 y + 0)/2 is Half this number
solve the equation y=4
the no. is 40

Let this 2-digit number be (10x+y ), where x, y both are single digit integer.

this number less than 6 times the ten digit of the number by 4:
Eq1: (10x+y) / 2 = 6x - 4
=> y = 2x - 8

Let the quotient is Q and R is the remainder, where Q, R are both integer
this number divisible by both 2 and 5 means :
(10x+y) divide by 2 or 5, remainder is 0
Eq2:
This number = (2*5) Q + R
=> this number = 10Q (R=0)
=> (10x + y) = 10Q
=> x + y/10 = Q
Since Q is integer, y must not be equal to 1 to 9. then y = 0

Put y = 0 into Eq1
X=4
So the number is 40

Let the ten digit be X, and the digit of the ones be y
the value of the number would then be (10x + y):

As the number is divisible by both 2 and 5,
the LCM of 2 and 5 is 10
so the unknown number must be divisible by 10
Hence, y = 0

(10x + y)/2 = 6x - 4
5x = 6x - 4
x = 4

So the 2-digit number is 40.