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If the first digit of a number is deleted, the number gets reduced by 5757 times. Find all such numbers.

Note by Swapnil Das
3 years, 11 months ago

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

Let p=a1a2a3an , q=a2a3anp=\overline{a_1a_2a_3 \dots a_n} \ , \ q=\overline{a_2a_3 \dots a_n} .

By given info. pq=57pq1=56pqq=5610n1a1q=56q=10n1a156=10n4a1×10356=10n4a1×537\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 qq is an integer , 77 is forced to divide a1a_1 which implies a1=7q=125×10n4a_1=7 \Rightarrow q = 125 \times 10^{n-4}.

Thus p=7×10n1+p=10n1+125×10n4=10n4(7000+125)=7125×10n4p=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 nn are number of digits of pp. So in general , these numbers are also in form 7125×10k kZ++{0}\boxed{7125 \times 10^k \ \forall k \in \mathbb{Z}^+ + \{0\}}.

Nihar Mahajan - 3 years, 11 months ago

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

Swapnil Das - 3 years, 11 months ago

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Is the number something 7171 \dots ?

Nihar Mahajan - 3 years, 11 months ago

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

Swapnil Das - 3 years, 11 months ago

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I have completed my solution , do review it :)

Nihar Mahajan - 3 years, 11 months ago

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Just multiply your answer by 10000

Swapnil Das - 3 years, 11 months ago

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No , not multiply. Just replace n4=kn-4 = k. Has this solution helped you?

Nihar Mahajan - 3 years, 11 months 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 - 3 years, 11 months ago

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

Nihar Mahajan - 3 years, 11 months ago

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Who said you are not good in Number Theory?

Aditya Chauhan - 3 years, 11 months 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 - 3 years, 11 months ago

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Let the number be a1a2a3......anˉ\bar { { a }_{ 1 }{ a }_{ 2 }{ a }_{ 3 }......{ a }_{ n } } The new number will be a2a3......anˉ\bar { { a }_{ 2 }{ a }_{ 3 }......{ a }_{ n } }

a1a2a3......anˉa2a3......anˉ=57Afterinproductforma1a2a3......anˉ=56\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.

Department 8 - 3 years, 11 months ago

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

Calvin Lin Staff - 3 years, 11 months ago

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

Department 8 - 3 years, 11 months ago

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@Department 8 Can you explain what is product form? Shouldn't the numerator be a1×10n1 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 - 3 years, 11 months ago

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

Department 8 - 3 years, 11 months ago

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Yes your numerator is incorrect. It is not true that a1a2a3ana2a3an=a1\overline{a_1a_2a_3\dots a_n} - \overline{a_2a_3\dots a_n} = a_1 . In fact it must be a1×10n1a_1 \times 10^{n-1} . Your solution implies that 1987654987654=11987654-987654=1 which is false and it must be 10000001000000 instead. BTW What is product form?

Nihar Mahajan - 3 years, 11 months ago

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Thanks, You gave the answer to the question yourself.

Department 8 - 3 years, 11 months ago

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Thanks for giving me the idea of subtracting 11 from both sides.

Nihar Mahajan - 3 years, 11 months ago

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