there is a number that is 5 times the sum of its digits. What is the number?

Question

tough one! who can get it!

Answer 1

45 is one such number.
0 is another.

One digit numbers:
a = 5a
4a = 0
a = 0
Answer = 0

Two digit numbers:
10a + b = 5(a + b)
10a + b = 5a + 5b
5a = 4b
Using digits (0 to 9) the only solution is a = 4, b = 5
Answer = 45

Three digit numbers:
100a + 10b + c = 5(a + b + c)
95a + 5b - 4c = 0
95a = 4c - 5b
4c - 5b can never be bigger than 36 (c = 9, b = 0), so there is no solution (a > 0) for a. This follows for all numbers with more than two digits.

The only answers are 0 and 45. I suspect the *intended* answer is 45:

(4 + 5) = 9
5 * 9 = 45

Answer 2

Not so tough, if you write the number as 10a + b.
Then (10a+b) = 5(a+b)
This, with the stipulation that a and b are whole numbers between 0 and 9, ought to do it, and since it's a multiple of 5, b = 5 or 0, ought to do it. Let's look at a final digit of 5, since that's more interesting.
(assuming it's a two digit number)
So 10a + 5 = 5a + 25
5a = 20 a = 5 b = 4, the number is 45
What if b = 0?
then 10a = 5a, so a = 0, so the number has to be 0. Not as interesting.
This little bit of algebra seems to indicate that the only solutions to your riddle, of one or two digits, is 0 and 45.
If it's a three digit number you'd write it as
100a + 10b + c = 5(a + b+ c)
Again, c would have to be 5 or 0.
If c = 0 then this reduces to 100a + 10b = 5a + 5b, which can be true only if a and b are 0 as well. So all the interesting solutions would be numbers ending in 5...
100a + 10b + 5 = 5(a + b + 5)
100a + 10b + 5 = 5a + 5b + 25
95a + 5b -20 = 0
So all your solutions would be of the form (a,b) = (x,y) whole numbers lying on the line 95a + 5b = 20
Since a and b are whole numbers, (nonnegative) it appears (unless I've goofed with the algebra) there are no solutios to your riddle for numbers with three digits. If you tried to do it for larger numbers, with four or more digits, you'd run into the same problem, only more so; so it appears that there are no solutions for numbers with three or more digits.

Answer 3

The remainder upon division by 9 will always be different once you multiply by 5, unless that remainder was zero; i.e. the only numbers that will work are divisible by 9. So I figured I'd try multiplying 9 by 5 and adding the digits...and I got lucky:

9x5 = 45; 4+5=9.

QED

Edit: Unfortunately, no other number will work. Any number that is a multiple of the sum of its digits is divisible by 9, and if it is divisible by 5 also, it is divisible by 45, the LCM of 9 and 5. But 90 doesn't work, and no three-digit number will work, as they have to be divisible by 9, and the sum of its digits will usually be 9 or 18, and 18 x 5 = 90, which has only two digits. The only way we can get a three digit number, after adding the digits of a multiple of 9 and multiplying by 5, is if we did 999, which gave us a digital root (or, the sum of its digits) of 27, and 27 x 5 = 135.

If there are four digits, it gets worse, as the number is at least the size of 1000, but the sum of its digits will always be less than 40, as they are three digits less than ten, and 40 x 5 = 200. Five-digit numbers are at least 10,000, and their digital roots are less than 50, and 50 x 5 = 250. The minimum number at this point will always be greater than its digital root times five, as the number keeps growing by a factor of 10, while the digital-root-times-five only increases by about the number 50 each time. So 45 is the only solution. QED

Answer 4

Well if we consider the case for 2 digit integral nos the we can say the no to be (10a+b) , a,b are single digit integers
A/Q 10a+b = 5 (a+b)
=> 5a = 4b

thus we can have the set :
a 0 4 8
b 0 5 10
No 0 45 90 and so on

But it is not satisfied for (8,10) since 10 is double digit no.

Thus only possible 2 two - digit nos are 0 , 45

Answer 5

People said 0 & 45 but I say the only possible answer is 45 since you mentioned sum of digit's' in your question. There should not be no other number.

Answer 6

let´s supose is of two digits a and b
10a+b=5(a+b) 5a=4b===> b/a=5/4 we can try with b=5 a=4
which gives the number 45 = 5(4+5)=45

Answer 7

45 and 0

4+5=9
9*5=45

0=0
0*5=0

Answer 8

45.

4 + 5 = 9 * 5 = 45

Answer 9

10a+b=5(a+b)
10a+b = 5a + 5b
5a=4b
a=(4/5)b
i wonder if we put in 15 for b if we'll get another answer
a = 12

120+15 = 135

135 = 5 (9) so it doesn't work unless the you have single digits.

Answer 10

4+5=9 and 9*5=45

I am sure there are other numbers that would work too...but I don't know what they are. Hehe