Question

Someone please help!
-it is a 4 digit number.
-the vaule of the digit in the thousands place is 10 times greater than the value of the digit in the hundreds place.
-the least digit is in the tens place.
-the sun of the digits is 19. -the digit in the ones place is 4 more than the digit in the hundreds place.
-whats the mystery number?

Answer 1

Answer:

5539

Step-by-step explanation:

You should conclude that the number in the thousands place and the one in the 100s place are the same.

x * 1000 + x * 100 + y + x + 4

So you could sum the digits as

3x + y + 4 = 19   Subtract 4 from both sides

3x +  y = 15

x can be six           That means y = 1 which is OK

x can be five          That means y = 3 which is OK

Now we run into a little grief. Can x be 4? No

x cannot be 4          that would mean that y = 19 - 12 = 7 The tens digit is too big.

Come to think of it, x cannot be 6 because then the units digit will be 10 -- not possible.

So the answer is

1000*5 + 100*5 + 10*3 + 9

5539 is the answer.

Thanks for posting.

Answer 2

We want to find a 4-digit number that meets some given conditions. We will find two, which are:

4438 and 5509

A general 4-digit number can be written as:

a*1000 + b*100 + c*10 + d

Where a, b, c, and d are single-digit numbers.

a is the thousands digit

b is the hundreds digit

c is the tens digit

d is the units digit.

Here we know that:

"the value of the digit in the thousands place is 10 times greater than the value of the digit in the hundreds place."

So the first term is 10 times the second term, this means that:

(a*1000) = 10*(b*100)

a*1000 = b*1000

a = b

-the least (the smaller one) digit is in the tens place.

so:

c < a, b, d

- The sun of the digits is 19.

So:

a + b + c+ d = 19

-The digit in the ones place is 4 more than the digit in the hundreds place.

so:

d = b + 4

Now we can rewrite all of these conditions as:

• b = a
• c < b, a, d
• a + b + c + d = 19
• d = b + 4

Using the first one, we can replace all the b's by a's in the other conditions, so we have:

• c <  a, d
• 2a + c + d = 19
• d = a + 4

Now we can use the third equation and replace it in the second one to get:

2a + c + d = 19

2a + c + (a + 4) = 19

3a + c + 4 = 19

3a + c = 19 - 4

3a + c = 15

Now we need to remember that:

c < a.

So we can try with different values of a.

For example, if a = 3, then:

3*3 + c = 15

9 + c = 15

c = 15 - 9 = 6

here we have c > a, so we can discard this.

If a = 4, then:

3*4 + c = 15

12 + c = 15

c = 15 - 12 = 3

c < a

From this we can construct our number:

a = b = 4

c = 3

d = b + 4 = 4 + 4 = 8

Then the number is:

4438

If a = 5, then:

3*5 + c = 15

15 + c = 15

c = 15 - 15 = 0

c < a

So we also can have:

a = b = 5

c = 0

d = b + 4 = 9

Such that the number is:

5509

And if a is equal to or larger than 6 we have problems, because

3*6 + c = 15

18 + c =15

c = 15 -18 = -3

And c can't be a be a negative number.

So we can conclude that we found two 4-digit numbers that meet all the conditions, and these are:

4438 and 5509

If you want to learn more, you can read:

https://brainly.com/question/19902993