An Unconventional NT Problem

Just a note: this is quite similar to the 2006 AIME I Problem #3. Remember to credit your sources!

Let's take a simple approach.

Let our number be $n$ , where $n$ is a positive integer of the form $\overline{abc}$ ( $a, b$ and $c$ do not necessarily need to be distinct) and that satisfies the above property. Thus, $\overline{bc}=\frac{\overline{abc}}{29}$ .

We know that the only possible other factor of $n$ must end in either $0$ or in $5$ , because those are the only numbers that, when multiplied by $29$ , produce a product with the same last digit. Thus, our search is limited to cases where the other factor is a multiple of $5$ . This makes our search a whole lot easier.

Now, we start to search for the number we seek. A quick listing of the numbers can give us $145, 290, 435, 580, 725$ -- STOP. That's it. We're done. $25 \times 29 = \boxed{725}.$

That's how I did it, any more efficient ways?

Instead of figuring out $n$ , let's figure out $\dfrac{n}{29}$ instead. Let's call $\dfrac{n}{29}=x$ . Clearly, $29x$ is at least a 3-digit number. Let's assume that is true. Thus, $29x$ can be represented by $100a+x$ , where $a$ is a digit. Since $29x=100a+x$ , we have $28x=100a$ , or $7x=25a$ . Therefore, $x=25$ and $a=7$ .

Finally, $n=29x=\boxed{725}$ and we are done.

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Wow, that solution was a lot less rigorous and a lot more wordy than I intended it to be!

Let consider the number that satisfies the condition is a two digit number like ab . Then $10a+b=29b$ so $b=\frac{5a}{14}$ It means the least possible value of a is 14, as b is an integer but it is not possible because a,b,c lies in the interval [0,9] and they are all integer. So the number is not a two digit number. Let consider the number is a three digit number like abc . Then $100a+10b+c=29(10b+c)$ so $10b+c=\frac{25c}{7}$ Then the least possible value of c is 7. So $10b+c=25$ . It means that b=2,c=5. Then the number is 725

let the leftmost digit be x.

then acc. to ques. , no. would be x * (power of 10) + n/29

i.e. n = x * (power of 10) + n/29

which implies, n - n/29 = x * (power of 10)

n*28/29 = x * (power of 10)

RHS is integral , therefore n must be a multiple of 29 say 29k.

so, 28*k = x * (power of 10)

4 7 k = x * (power of 10) , there is no no. which when multiplied by 7 fetches a power of 10,
so x is essentially a multiple of 7 ,infact it is 7 as x is a digit.

now, 4*k = power of 10

clearly min. k = 25 , which gives n= 29 * 25 = 725

lets say number is 100x+10y+z

so equality will become

10y+z = 100x+10y+z / 29

28(10y+z) = 100x

4*7(10y+z) = 100 x

x has to be 7

4*(10y+z) = 100 = 4 * 25

and y = 2 and z = 5

so number 725

Let the first digit be $a$ . If $n$ has (b+1) digits, removing $a$ would reduce its value by $a \times 10^b$

So we have $n - a \times 10^b = \frac{n}{29}$

$\frac{28n}{29} = a \times 10^b$

$28n = 29 a \times 10^b$

Since 28 and 29 are coprime, $a \times 10^b$ must be divisible by $28$ or $2^2 \times 7$ .

$10^b$ is no way divisible by 7 so $b$ must be divisible by 7. Since $b$ is single digit, $b=7$

Now $28n = 29 \times 7 \times 10^b$

$4n = 29 \times 10^b$

For $29 \times 10^b$ to be a multiple of 4, $b$ must be at least 2

$4n = 2900$

$n = \boxed{725}$