# Need help

If the first digit of a number is deleted, the number gets reduced by $57$ times. Find all such numbers.

Note by Swapnil Das
4 years, 5 months ago

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I am not that good in NT but still this is my try:

Let $p=\overline{a_1a_2a_3 \dots a_n} \ , \ q=\overline{a_2a_3 \dots a_n}$ .

By given info. $\dfrac{p}{q}=57 \Rightarrow \dfrac{p}{q}-1=56 \Rightarrow \dfrac{p-q}{q}=56 \\ \Rightarrow \dfrac{10^{n-1}a_1}{q} = 56 \Rightarrow q = \dfrac{10^{n-1}a_1}{56} = \dfrac{10^{n-4}a_1 \times 10^3}{56} = \dfrac{10^{n-4}a_1 \times 5^3}{7}$.

Since $q$ is an integer , $7$ is forced to divide $a_1$ which implies $a_1=7 \Rightarrow q = 125 \times 10^{n-4}$.

Thus $p=7\times 10^{n-1}+p = 10^{n-1}+125 \times 10^{n-4} = 10^{n-4}(7000+125) = \boxed{7125 \times 10^{n-4}}$ where $n$ are number of digits of $p$. So in general , these numbers are also in form $\boxed{7125 \times 10^k \ \forall k \in \mathbb{Z}^+ + \{0\}}$.

- 4 years, 5 months ago

Nah bro, its at least not in accordance to the answer key :(

- 4 years, 5 months ago

Is the number something $71 \dots$ ?

- 4 years, 5 months ago

Yeah, but not fully. The number has a general form.

- 4 years, 5 months ago

I have completed my solution , do review it :)

- 4 years, 5 months ago

- 4 years, 5 months ago

No , not multiply. Just replace $n-4 = k$. Has this solution helped you?

- 4 years, 5 months ago

Aahh, Yes.I get it. It has really helped. Question to you: Is this a good problem?

- 4 years, 5 months ago

Yes , it is a very nice problem. Keep sharing more!

- 4 years, 5 months ago

Who said you are not good in Number Theory?

- 4 years, 5 months ago

Me. I can understand my strong/weak points better than anyone else.(This is actually applicable to all)

- 4 years, 5 months ago

Let the number be $\bar { { a }_{ 1 }{ a }_{ 2 }{ a }_{ 3 }......{ a }_{ n } }$ The new number will be $\bar { { a }_{ 2 }{ a }_{ 3 }......{ a }_{ n } }$

$\frac{\bar { { a }_{ 1 }{ a }_{ 2 }{ a }_{ 3 }......{ a }_{ n } } }{\bar { { a }_{ 2 }{ a }_{ 3 }......{ a }_{ n } } }=57 \\ \quad After \quad in \quad product \quad form \\ \frac{{a}_{1}}{\bar { { a }_{ 2 }{ a }_{ 3 }......{ a }_{ n } } }=56$

Now by keeping factors try to find out the numbers.

- 4 years, 5 months ago

Hm, in the second equation, are you missing a lot of 0's aren't you?

Staff - 4 years, 5 months ago

By applying product form of a number and see the answer will be the same.

- 4 years, 5 months ago

Can you explain what is product form? Shouldn't the numerator be $a_1 \times 10^{n-1}$, since you've only subtracted 1 from both sides?

Can you explain what "now by keeping factors" mean?

Staff - 4 years, 5 months ago

That's what I was talking about.

- 4 years, 5 months ago

Yes your numerator is incorrect. It is not true that $\overline{a_1a_2a_3\dots a_n} - \overline{a_2a_3\dots a_n} = a_1$ . In fact it must be $a_1 \times 10^{n-1}$ . Your solution implies that $1987654-987654=1$ which is false and it must be $1000000$ instead. BTW What is product form?

- 4 years, 5 months ago

Thanks, You gave the answer to the question yourself.

- 4 years, 5 months ago

Thanks for giving me the idea of subtracting $1$ from both sides.

- 4 years, 5 months ago