Question

In a four digit number having non zero and distinct digits,the sum of the digits at the units position and the tens position is equal to the sum of the remaining two digits. the sum of the digits at tens and the hundreds position is three times the sum of the remaining two digits ,if the sum of the digits is not more than 20, then how many such four digit no. are possible?

Answer 1

ay the number is abcd 
Since b+c = 3(a+d), the sum of all the numbers is 4(a+d) <=20 
So a+d <= 5. This is a pretty big restriction. That's what we are looking for, some restriction that cuts down the cases dramatically. 

Also, a and d are different. 

So a and d can only be (in either order): 
1 2 (then b+c = 9) 
1 3 (then b+c = 12) 
1 4 (then b+c = 15) 
2 3 (then b+c = 15) 

Now a + b = d + c, so let's look at the differences between a and d 
If a=1 and d=2, then b is 1 greater than c, so b=5 and = 4 
If a=2 and d=1, then b=4 and c=5 

If a=1 and d=3, then b is 2 greater than c, so b=7 and c=5 
if a=3 and d=1, then b = 5 and c=7 

if a=1 and d=4, then b is 3 greater than c, so b=9 and c =6 
if a=4 and d=1, then b=6 and c=9 

If a=2 and d=3, then b is 1 greater than c, so b=8 and c=7 
if a=3 and d=2, then b= 7 and c=8 

Those are the only 8 solutions to all the conditions.