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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, 1 month 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 \(\boxed{7125 \times 10^k \ \forall k \in \mathbb{Z}^+ + \{0\}}\).

Nihar Mahajan - 2 years, 1 month ago

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Who said you are not good in Number Theory?

Aditya Chauhan - 2 years, 1 month ago

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

Nihar Mahajan - 2 years, 1 month ago

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Just multiply your answer by 10000

Swapnil Das - 2 years, 1 month ago

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

Nihar Mahajan - 2 years, 1 month ago

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

Swapnil Das - 2 years, 1 month ago

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

Nihar Mahajan - 2 years, 1 month ago

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

Swapnil Das - 2 years, 1 month ago

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Is the number something \(71 \dots\) ?

Nihar Mahajan - 2 years, 1 month ago

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

Swapnil Das - 2 years, 1 month ago

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I have completed my solution , do review it :)

Nihar Mahajan - 2 years, 1 month 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.

Lakshya Sinha - 2 years, 1 month 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?

Nihar Mahajan - 2 years, 1 month ago

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Thanks, You gave the answer to the question yourself.

Lakshya Sinha - 2 years, 1 month ago

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

Calvin Lin Staff - 2 years, 1 month ago

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

Lakshya Sinha - 2 years, 1 month ago

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@Lakshya Sinha 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?

Calvin Lin Staff - 2 years, 1 month ago

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@Calvin Lin That's what I was talking about.

Lakshya Sinha - 2 years, 1 month ago

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Thanks for giving me the idea of subtracting \(1\) from both sides.

Nihar Mahajan - 2 years, 1 month ago

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