The Special

Call the digits from right to left $a,b,c,\cdots \in \{0,1,\dots,9\}$ . We have $5(a + b + c + d + \cdots) = a + 10b + 100c + 1000d + \cdots.$ $4a = 5b + 95c + 995d + \cdots$ If any of $c,d,\dots$ were non-zero, $4a$ should be at least 95, which is impossible. Therefore the solution must have two digits at most. $4a = 5b$ The only non-zero solution satisfying the constraints on $a$ and $b$ is $a = 5,\ b = 4:\ \ \ \ \ \ 5(5 + 4) = 5 + 40 = \boxed{45}.$

1. Clearly, for a single digit, this property doesn't exist.

2.Checking for a $2$ digit number $\overline{ab}$ , we get $(a+b)\times 5 = 10a+b=5a+5b$ $\implies \frac{a}{b} =\frac45$ . Clearly, it is only possible if number is $45$ .

3.Checking for a $3$ digit number $\overline{abc}$ , we get $(a+b+c)\times 5=100a+10b+c \implies 4c=95a+5b$ . Now, for $a=1$ and $b=0$ $\color{#3D99F6} \text{(the minimum case)}$ $c$ is not an integer and even if $a=b=1$ , $c$ has to be equal to $25$ , but its not possible since $c$ is a single digit. Therefore, a $3$ digit number also doesn't holds this property.

And now for subsequent values of $a,b,c$ ,we will not get any single digit number because they will become larger and larger. Therefore, no positive number other than $45$ exhibits this property.

(4+5)(5) = 45

(3+2)(5) does not equal 30.

Let $x=\overline{ab}$

Then we can write, $x=\overline{ab}=10a+b$

From the question we can write, $5\times(a+b)=x=10a+b\\5a+5b=10a+b\\5a=4b\\a=\frac{4b}{5}$

Here $a$ would be a integer if $\frac{4b}{5}$ is a intiger and then we can say for $a$ being integer $b$ must be divisible by $5$

Hence $b$ can be $=0,5,10,15,...$

If $b=0$ then $a$ comes $0$ . $\color{#D61F06}{\text{It's not possible.}}$

Again $b=5$ then $a$ comes $4$ $\color{#20A900}{\text{It's possible.}}$

Again $b\neq10,15....$ because $b$ is a single digit.

So, we can get only one answer, which is $x=\boxed{45}$

(X+Y)5=10X+Y

5X+5Y=10X+Y

5Y-Y=10X-5X

4Y=5X

4Y/5=X

The only single digit positive whole number Y that fits this equation with a result that is also a single digit positive number is Y=5 making X=4.

Oh it is simple we can represent 2 numbers of x as a_b. a+b*5 = x a+b = x/5 thus meaning x is dividable by 5 and has 2 numbers in it so x can be {10, 15, 20, 25, 30, 35, 40, 45...} I mean i just solved it like 1+0 = 10/5 => 1=2 only answer I got to was 45 :D and that was only one working... I mean I am not genius so...

How is zero not a solution?

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# -*- coding: utf-8 -*-
"""
Created on Tue Nov 28 18:27:20 2017

@author: Michael Fitzgerald
"""
for i in range(10,1000):
    digits = [int(x) for x in str(i)]
    ans = sum(digits)*5
    if ans == i:
        print i

45

5(x+y)=10x+y 5x=4y

I can prove this by elimination.

Step 1: All digits of a number are positive integers and therefore must add up to a positive integer. Any number that is not a multiple of five, if divided by five, will not return a positive integer; therefore, the number must be divisible by five.

Step 2: Three-digit numbers are out. The highest sum of digits any three-digit number divisible by five can have is 23; 23 * 5 is 115. Any number between 100 and 195 is also out because 1 + 9 + 5, the maximum, = 15 * 5 = 75 which is less than 100.

Step 3: By this same logic, each further order of magnitude is out, because the maximum sum of digits * 5 increases by 45 each order of magnitude, while the actual number you need increases by, well, orders of magnitude.

Step 4: Odd multiples of five whose digits add to an even number are out because even numbers * 5 end in 0. Even multiples of five whose digits add to an odd number are out because odd numbers * 5 end in 5. (Proof: an odd * an odd is an odd; an even times an odd is an even; multiples of five can only end in one odd number or one even number; therefore, an even multiple of five always ends in 0 and an odd multiple of five always ends in five.)

Step 5: Test the remaining cases. The digits of 20, 40, 60, and 80 all multiply to half of their source. 5, 25, 45, 65, and 85 remain; 5, 25, 65, and 85 do not resolve properly; 45 is the only answer.

Hello, I am new to this. What I did was I used a formula I made up, $digit(n, i, base)=\lfloor{}n÷base^{i}\rfloor{} \bmod{} base$ where n is the number, i is the index, and base is the base. I summed up the digits then all the digits from index 0 to index 20 (because I figured beyond that was unnecessary and I was too lazy to type $\lfloor{}\log_{base}{x}\rfloor{}+1$ the formula to sum up the digits of x was $\sum\limits_{i=0}^{20} digit(x, i, base)$

I checked it on a graphing calculator. 45 looked right. I did the math. It was right. I am not sure I could prove this is the only solution with my method, though.

Btw assume we all have 10 fingers and count in base-10.

The multiples of 9 satiefies the property that adding all the digits then multiplied by a constant equals itself. This is always true because the rule of divisibility of 9 tells us that any multiple of 9, when added all of its digits, and if you have two or more digits, repeat, you always get 9. In this question, the constant is 5. Therefore, 9*5=45, which is the only answer.

I solved it like this:

$10 \cdot a + 1 \cdot b = 5(a+b)$

$10a + b = 5a+5b$

Subtract -5a and -b both sides

$5a = 4b$

We see that b > a and we need to find a and b such that there is an equality

We need to try for different numbers, lets try with b.

if b = 1 then a is a fraction -> not valid because we need an integer

b=2 then a is a fraction -> not valid because we need an integer

b=3 then a is a fraction -> not valid because we need an integer

b=4 then a is a fraction -> not valid because we need an integer

b=5 then a is 5a=20, --> a=4 NOW WE HAVE AN INTEGER

Test: $a=4, b=5$

$10 \cdot 4 + 5 = 5(4+5)$

$45 = 5 \cdot 9$

$45 = 45$

Success

Let unit digit is p, Tens digit is q,

Then No. is (10q+p), Acc. to Question. (q+p)5= 10q+p, 4p = 5q, q = 4p/5,

Since p & q both are positive integer so when q =4, then p=5, So, number is 40+5=45

(x+y) * 5 = 10x + y will eventually give 4/5y = x from there I guessed

There are actually two solutions. The second one is zero. Sum of digits of 0: 0. 0x5=0.

Shame! The name of the problem gives the solution away!

The expansion of our number $N$ in base $10$ is a sum of the form $a_{0} + a_{1}*10 + a_{2}*10^2 + ... + a_{n-1}*10^{n-1}$ ; this is a number that has $n$ digits.

Since all digits of the form $a_{i}$ can be at most equal to $9$ , the sum of all digits $\sum_{i=0}^{n-1} a_{i}$ must be less than or equal to $9n$ .Thus, $\sum_{i=0}^{n-1} a_{i} \leq 9n$ , and therefore $5\sum_{i=0}^{n-1} a_{i} \leq 5*9n = 45n$ , from which we can conclude that $N \leq 45n$ .

If we take the logarithm base 10 on both sides, we get $log(N) \leq log(45n) = log(45) + log(n)$ . Since $N$ has $n$ digits, $n - 1 \leq log(N) \leq n$ , and thus $\lfloor log(N) \rfloor = n - 1$ ; what this means is that given $\lfloor log(N) \rfloor \leq \lfloor log(45) + log(n) \rfloor$ , we have that $n - 1 \leq \lfloor log(45) + log(n) \rfloor$

If we plug in values for $n$ , we see that for any value $n \geq 3$ , the quantity $n - 1$ surpasses $\lfloor log(45) + log(n) \rfloor$ , and thus our number $N$ can have at most 2 digits.

If $n$ is a single digit number, $n$ is equal to $a_{0}$ , and we'd have $a_{0} = 5*a_{0}$ and in turn $a_{0} = 0$ , which would make $N = 0$ and this is not a valid solution.

If $n$ has two digits, $n = a_{0} + a_{1}*10$ , and then $a_{0} + 10*a_{1} = 5(a_{0} + a_{1})$ ; this gives us $4a_{0} = 5a_{1}$ , and since both are numbers between $0$ and $9$ (inclusive, except $a_{0}$ cannot be $0$ if $a_{1}$ is also $0$ ), the only numbers that satisfy all conditions are $a_{0} = 5$ and a_[1} = 4 , which in turn means that $N = 45$ , which is the solution sought.

There is only one because we can see how if the number increase the value of the answer with the number x is at a greater difference and the same for a lo were number... That can be explained with a graph and a parabòlic line

sum of digital is multimplied by 5.surely last digit of required no =0/5 so (a+b)×5=10a+b $a÷b=4÷5 $(ab)=(45)

The answer must satisfy these:

1. The number must be 2-digits so it can have a sum of its digits.

2. The number must have a factor of five. (Because we have to times five later on, to have a chance of being the same, the number has to have a factor of five.)

Then we can start calculating: 10, 15, 20, 25, 30, 35, 40, 45....

The answer is 45 because (4+5)x5=9x5=45

a = tenth digit

b = oneth digit

(a + b) x 5 = 10a + b

5a + 5b = 10a + b

4b = 5a

if a = 4 and b = 5

substitute in 20 = 20 is correct

anwser is 45

One solution would be writing multiples of 5, adding their digits an then multiplying them by 5. Something like this:

$05 = (0+5) = 5 * 5 = 25$

$10 = (1+0) = 1 * 5 = 5$

$15 = (1+5) = 6 * 5 = 30$

$20 = (2+0) = 2 * 5 = 10$

$25 = (2+5) = 7 * 5 = 35$

$30 = (3+0) = 3 * 5 = 15$

$35 = (3+5) = 8 * 5 = 40$

$40 = (4+0) = 4 * 5 = 20$

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

$50 = (5+0) = 5 * 5 = 25$

$.$

$.$

$.$

X=xy (x+y)×5=x X+y=x/5 Y=x/5-x Y=-4x/5 5y=4x use only absolute values since the number is xy then the number is 45. (4+5)×5=9×5=45

• Let's begin assuming that $x$ is an $n$ -digit positive integer.

• Then, $x = 10^{n-1}a_{n-1}+ 10^{n-2}a_{n-2}+ \ldots \ldots \ldots + 10^3a_3+10^2a_2+10a_1+a_0$ .

• And sum of digits of $x = a_0+a_1+a_2+a_3+ \ldots \ldots \ldots + a_{n-1}$ .

• The problem statement demands $x$ must satisfy the following equality.

$10^{n-1}a_{n-1}+ 10^{n-2}a_{n-2}+ \ldots \ldots \ldots + 10^3a_3+10^2a_2+10a_1+a_0 = 5 \times (a_0+a_1+a_2+a_3+ \ldots \ldots \ldots + a_{n-1})$

$\Leftrightarrow (10^{n-1}-5)a_{n-1}+ (10^{n-2}-5)a_{n-2}+ \ldots \ldots \ldots + (10^3-5)a_3+(10^2-5)a_2+(10-5)a_1 = 4a_0 \ldots \ldots \ldots (1)$

• Now, we notice, under the restriction that $n\geq 3$ , the minimum value of the left side expression of $(1)$ occurs when $a_{n-1}=a_{n-2}=\ldots \ldots \ldots = a_4=a_3=0$ , $a_2=1$ and $a_1=0$ , which is $95$ . On the other hand, under the very same restriction, the maximum value of the right side expression of $(1)$ occurs when $a_0=9$ , which is $36$ . As for $n\geq 3$ , the minimum value of the left side expression of $(1)$ exceeds even the maximum value of the right side expression of $(1)$ , no positive integer $x$ with $n\geq 3$ can satisfy $(1)$ .

• So, $x$ is, at most, a $2$ -digit positive integer, with $x= 10a_1+a_0$ .

• We have, from $(1)$ , $5a_1 = 4a_0 \ldots \ldots \ldots (2)$ .

• Now, we see, $x$ cannot be a single-digit number; because $x$ is a single-digit number $\implies a_1 = 0 \implies a_0 = 0 \implies x = 0$ , which contradicts the fact that $x$ is a positive integer.

• So, $x$ is a $2$ -digit positive integer; and none of $a_1$ and $a_0$ equals $0$ .

• In accordance with $(2)$ , $4 | a_1$ and $5 | a_0$ . So, the set of candidates for $a_1$ is $\{4,8\}$ and for $a_0$ is $\{5\}$ . We are left with two candidates for $x$ , namely $45$ and $85$ .

• It's easy to see $85$ doesn't qualify, but $\boxed{45}$ does. And uniqueness of $45$ is proved too.

10x+y=5(x+y) ---- equation 1 5x=4y 5x/4=y Since we need y to be integer, x must be a multiple of 4. But, 4 itself is a multiple of 4. Therefore,for x=4, y=5*4/4=5. By substituting in equation 1, that x=4 and y=5, The number is 45. It is the only such number because on plugging x=8,12,16,...(the next multiples of 4), y becomes 10 which is not allowed because only the numbers 0 to 9 are allowed to take place of y as according to the base 10 number system.