a 3 digit number my tens digit is 5 more than my ones digitmy hundreds digit is eight less than my tens find?

Question

i am a

Answer 1

Let
h = hundreds digit
t = tens digit
u = units digit

we know that
t = u + 5 (since the tens digit is 5 more than the ones digit)
h = t - 8 (since the hundreds digit is 8 less than the tens digit)

Let us take a look at the second equation:
h = t - 8

Since h and t are digits of a number, then we know that h and t are single digits.

Now, h and t cannot be negative, so the only possible value for t is 9 or 8 (since 10 is no longer a digit and any value less than 8 will make h negative).

Thus, if t = 9, h = 9 - 8 = 1, and
if t = 8, h = 8 - 8 = 0.

But since we know that h is the hundreds digit, h cannot be equal to zero. Therefore, h = 1. And also t = 9. Now, using the first equation,
t = u + 5

We can solve for u
u = t - 5

Since t = 9, then
u = 9 - 5

Therefore,
u = 4

Now, we know the values
u = 4,
t = 9, and
h = 1.

Therefore, the 3-digit number is
194.

^_^

Answer 2

Well, if the hundreds digit is 8 less than your tens digit, there is only one possibility: the hundreds digit must be 1 (because if it were 0, it wouldn't be a 3-digit number), and the tens digit must be 9. So the ones digit must be 4, and the number must be 194.

Answer 3

The digits are [x-3][x+5][x]
Since the first digit has to be >=1, and the second digit has to be <=9, the only possibility is x=4:
194.

Answer 4

enable the type be xyz. From the given fact: I deduce the reality under. (a million) y = z + 5 (2) x = y - 8 (3) 0 = < y,z < 10 (because of the fact they are single digit each and each) (4) 0 < x < 10 (so far the numbers is a three digit, subsequently it extremely is a non - 0 numbers) From (a million), (2), we get: y = z + 5 ........ (a million) x = y - 8 ......... (2) From (2) x = y - 8 y = x + 8 ............. equation(5) by utilising equating equation (a million) and equation (5) y = y z + 5 = x + 8 x = z + 5 - 8 x = z - 3 ........................ equation (6) From (4), (6): z - 3 > 0 ? z > 3 subsequently, z > 3 From (a million), (3): z + 5 < 10 z = < 10 - 5 ? z < 5 when you consider that 3 < z < 5, subsequently z = 4. utilising (a million), (2), we get: y = 4 + 5 = 9 x = 9 - 8 = a million ? x = a million, y = 9, z = 4 because of the fact the type is xyz subsequently the type is 194.

Answer 5

stephen is right!

Another way is:

let ones digit = a
let tens digit = b
let hundredths digit = c

Our number is therefore equal to 100c + 10b + a

also, given:
b = a + 5
c = b - 8

plugging these into our main number equation:
100(b-8) + 10b + (b-5)
or,
111b - 805

this is the number we want, we just have to plug in our own values for 'b', but make sure that it is a three digit number. By doing a little bit of work, we get b=10. Therefore our number is 194, as stephen mentioned.

Answer 6

194