Question

31. In a four-digit number, the sum of the middle
two digits is twice the units digit. The sum of the
hundreds digit and six times the thousands digit is
twice the sum of the other two digits. The sum of the
units digit and five times the thousands digit is twice
the hundreds digit. How many values can the
four-digit number assume?

Answer 1

Answer:

Let the four digits be a, b, c, d

Given

b + c = 2d

b + 6a = 2(c + d)

d + 5a = 2b

Write b, c, d in form multiples of a

d = 2b - 5a

b + c = 2d

b + c = 2(2b-5a)

b + c = 4b - 10a

3b = 10a + c

b + 6a = 2c + 2d

b + 6a = 2c + 4b - 10a

3b = 16a - 2c

Therefore,

10a + c = 16a - 2c

6a = 3c

c = 2a

3b = 16a - 2c = 16a - 2(2a) = 16a - 4a = 12a

b = 4a

d = 2b - 5a = 2(4a) - 5a = 3a

d = 3a

Now possibilities are

If a = 1 then b = 4, c = 2, d = 3

If a = 2 then b = 8, c = 4, d = 6

If a = 3 then b = 16 which is not possible because a,b,c,d are single digit numbers

Therefore, The number can Assume only 2 possibilities ( 1423, 2846 )

Answer 2

Answer:

2

Step-by-step explanation:

Given that,

b + c = 2d → (1)

b + 6a = 2(c + d) → (2)

d + 5a = 2b → (3)

Let us that the equations (1), (2) and (3) as the linear

equations in a, b and c and express the values of a, b and c

in terms of b.

By (2) − (1), we get 3c = 6a ⇒ a =

2

c

By substituting c = 2a in (1), it becomes

2d − 2a = b → (4)

Subtracting (4) from 2 × (3), we get

12a = 3b ⇒ a =

4

b

As c = 2a, c =

2

b

By substituting a =

4

b

in (4), we get 2d =

2

b3 ⇒ d =

4

b3

∴ a : b : c : d = 1 : 4 : 2 : 3

∴ ‘abcd’ can be 1423 or 2846