Question

3. In a four-digit number, the sum of the digits in the
units and the tens places is equal to the sum of the
other two digits. The sum of the digits in the tens
and the hundreds places is twice the sum of the
other two digits. If the sum of the digits of the
number is more than 20, then the digit in the units
place can be
(A) 6 (B) 7 (C) 8 (0) 5​

Answer 1

Answer: unit digit will be 5

Step-by-step explanation:

Let the number be abcd

We have a+b = c+d and b+c = 2(a+d) and a+b+c+d

a+b+c+d = 2(a+b).......1

a+b+c+d = 3(a+d).....2

From 1 and 2 we can say that digit sum is a multiple of 6

so the digit sum can be 24,30,36

36 is not possible as it implies all digits 9 which contradicts b+c = 2(a+d)

if digit sum is 30 then,

a+b=c+d=15

and a+d = 10 and b+c = 20 No solution as one digit can be 9 maximum.

Hence 24 is the digit sum

Now,

a+b = c+d = 12

b+c = 16 , a + d = 8

from a+d = 8 ;

(a,d) can be (4,4) ,(5,3),(3,5)

which implies

(b,c) = (8,8),(7,9),(9,7)

(6,2) pair of (a,d) is rejected as it will make b= 10 which is not possible

Therefore there are 3 values for unit digit = 3,4,5

Since only 5 is in option so this our answer.