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

Note by Swapnil Das
2 years, 7 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 $$\boxed{7125 \times 10^k \ \forall k \in \mathbb{Z}^+ + \{0\}}$$.

- 2 years, 7 months ago

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Who said you are not good in Number Theory?

- 2 years, 7 months ago

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

- 2 years, 7 months ago

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Just multiply your answer by 10000

- 2 years, 7 months ago

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

- 2 years, 7 months ago

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

- 2 years, 7 months ago

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

- 2 years, 7 months ago

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

- 2 years, 7 months ago

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Is the number something $$71 \dots$$ ?

- 2 years, 7 months ago

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

- 2 years, 7 months ago

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I have completed my solution , do review it :)

- 2 years, 7 months ago

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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.

- 2 years, 7 months ago

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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?

- 2 years, 7 months ago

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Thanks, You gave the answer to the question yourself.

- 2 years, 7 months ago

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

Staff - 2 years, 7 months ago

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

- 2 years, 7 months ago

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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 - 2 years, 7 months ago

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That's what I was talking about.

- 2 years, 7 months ago

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

- 2 years, 7 months ago

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