A 4 digit number, having all digits different and non odd, has sum of all digits equal to 12. The sum of middle two digits is equal to the first digit. Second digit is greater than third digit. The product of first 3 digits will be
Let the number is 1000x + 100y + 10z + p
According to the given conditions
x + y + z + p = 12 ..................(1)
y + z = x ..................(2)
Using y + z from (2) into (1)
we get 2x + p = 12
Possible combinations for the values of x and p are given below
x = 2, p = 8 ...................(A)
x = 4, p = 4 ...................(B)
x = 6, p = 0 ...................(C)
If (A) is correct then by (2) either y or z will become equal to x
If (B) is correct then by (2) two cases arise
(1) either y or z will become equal to x
(2) or y will become equal to z
so x = 6, p = 0 will be the case
so y + z = 6
=> y = 4 and z = 2
so number is 6420.