Question

My thousands digit is three times my hundreds digit. the sum of my tens digit and one digit is one less than my thousandsdigit. my hundreds digit is two less than my tens digit. what 4-digit number am i?

Answer 1

Answer:

6241 is one answer , but see below.

Step-by-step explanation:

If the number is vwxy we have:

v = 3w

x + y = v - 1

w = x - 2

From equations 1 and 2:

v = 3(x - 2).

The only possible values of v are 3, 6 or 9.

If v = 3:-  3 =  3(3 - 2)  so x = 3 also.

If v = 6:-   6 =  3(4 - 2) so x = 4.

If v = 9:-   9 = 3(5 - 2) so x = 5.

Now consider the first equation x + y = v - 1:-

If v = 3 then 3 + y = 3 - 1  giving y = -1 so v = 3 is not possible.

If v = 6  then 4 + y =  6 - 1  giving y =  1  which is a possible value.

If v = 9 then 5 + y = 9 - 1 giving y = 3 which is also possible.

If  x = 4 w = 4 - 2 = 2

If x = 5 w = 5-2 = 3.

Thus the possible values are   6241 or 9353.

If all digits must be different the answer is 6241.

Answer 2

The four digit number which we're considering here is 6241 or 9353

How can we express numbers belonging to decimal numeral system?

Suppose that the number is abcd.efg and belongs to the decimal numerical system.

Remember one thing that the place on the left of decimal point is called ones.
Move one-one places to right, and divide by 10 and 10 more and more.

Move one-one places to left, and multiply (un-divide) by 10  and 10 more and more.

Thus, we write abcd.efg as:

$a \times 10^3 + b \times 10^2 + c \times 10^1 + d \times 10^{0} + e \times 10^{-1} + f \times 10^{-2} + g \times 10^{-3}$

We will call d as ones digit, c as tens digit (it get multiplied by 10), b as hundreds digit etc.

We call e as digit at 10th place (division by 10), f as digit at hundredth place and so on.

Let the four digit number be abcd,

then we get:

$a \times 10^3 + b \times 10^2 + c \times 10^1 + d \times 10^{0}$

1. Thousands digit = a
2. Hundreds digit = b
3. Tens digit = c
4. Ones digit = d

Now, it is given that:

• Thousands digit is 3 times hundreds digit, or: $a = 3 \times b$
• The sum of tens digit and ones digit is 1 less than thousands digit, or: $c+d = a - 1$

• The hundreds digit is 2 less than tens digit, or:  $b = c - 2$

Expressing every digit in terms of a:

a = 3b or b = a/3

Since b = c-2, therefore c = b+2

Thus, as c+d = a -1, therefore b+2+d = a-1

Since a = 3b, therefore b+2+d = 3b - 1 or 3+d = 2b or b = (d+3)/2

Since a = 3b, therefore a = 3(d+3)/2

Since  c+d = a-1, therefore c + d = 3(d+3)/2 - 1, or

$2c + 2d = 3d + 9 - 2 \\\\c = \dfrac{d+7}{2}$

Since a = 3(d+3)/2, therefore d = 2a/3 - 3,

and therefore $c = \dfrac{2a/3 - 3 + 7}{2} =a/3 + 2$
So we got:

• a = a
• b = a/3
• c = a/3 + 2
• d = 2a/3 - 3

Thus, the digit is:

$a \times 10^3 + b \times 10^2 + c \times 10^1 + d \times 10^{0}\\\\a \times 10^3 + \dfrac{a}{3}\times 10^2 + (\dfrac{a}{3} + 2) \times 10 + (\dfrac{2a}{3} - 3)$

Each digit of a number is integer between 0 and 9 (0 and 9 included).

a cannot be 0 since its a four digit number, and a is leftmost digit.

That means, b = a/3  needs to be integer.

The numbers from 1 to 9 which are divisible by 3 are 3,6 or9

Thus, a = 3, or 6 or 9

If a = 3, then d = 2a/3 - 3 = 2-3 = -1 (a negative number), which isn't possible.

So a = 6 or 9.

For a = 6, we get the four digit number as:

$a \times 10^3 + \dfrac{a}{3}\times 10^2 + (\dfrac{a}{3} + 2) \times 10 + (\dfrac{2a}{3} - 3)\\\\or\\\\6 \times 10^3 + 2 \times 10^2 + (2+2) \times 10 + (4 -3)\\\\or\\\\6000 + 200 + 40 + 1 = 6241$

For a = 9, we get the four digit number as:

$a \times 10^3 + \dfrac{a}{3}\times 10^2 + (\dfrac{a}{3} + 2) \times 10 + (\dfrac{2a}{3} - 3)\\\\or\\\\9 \times 10^3 + 3 \times 10^2 + (3+2) \times 10 + (6 -3)\\\\or\\\\9000 + 300 + 50 + 3 = 9353$

Thus, the four digit number which we're considering here is 6241 or 9353

Learn more about place value here:

https://brainly.com/question/17983725