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
·
1 year, 1 month ago

@Aditya Chauhan
–
Me. I can understand my strong/weak points better than anyone else.(This is actually applicable to all)
–
Nihar Mahajan
·
1 year, 1 month 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.
–
Lakshya Sinha
·
1 year, 1 month ago

Log in to reply

@Lakshya Sinha
–
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
·
1 year, 1 month ago

@Lakshya Sinha
–
Hm, in the second equation, are you missing a lot of 0's aren't you?
–
Calvin Lin
Staff
·
1 year, 1 month ago

Log in to reply

@Calvin Lin
–
By applying product form of a number and see the answer will be the same.
–
Lakshya Sinha
·
1 year, 1 month ago

Log in to reply

@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
·
1 year, 1 month ago

## Comments

Sort by:

TopNewestI 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 · 1 year, 1 month ago

Log in to reply

– Aditya Chauhan · 1 year, 1 month ago

Who said you are not good in Number Theory?Log in to reply

– Nihar Mahajan · 1 year, 1 month ago

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

– Swapnil Das · 1 year, 1 month ago

Just multiply your answer by 10000Log in to reply

– Nihar Mahajan · 1 year, 1 month ago

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

– Swapnil Das · 1 year, 1 month ago

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

– Nihar Mahajan · 1 year, 1 month ago

Yes , it is a very nice problem. Keep sharing more!Log in to reply

– Swapnil Das · 1 year, 1 month ago

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

– Nihar Mahajan · 1 year, 1 month ago

Is the number something \(71 \dots\) ?Log in to reply

– Swapnil Das · 1 year, 1 month ago

Yeah, but not fully. The number has a general form.Log in to reply

– Nihar Mahajan · 1 year, 1 month ago

I have completed my solution , do review it :)Log in to reply

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 · 1 year, 1 month ago

Log in to reply

– Nihar Mahajan · 1 year, 1 month 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?Log in to reply

– Lakshya Sinha · 1 year, 1 month ago

Thanks, You gave the answer to the question yourself.Log in to reply

– Calvin Lin Staff · 1 year, 1 month ago

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

– Lakshya Sinha · 1 year, 1 month ago

By applying product form of a number and see the answer will be the same.Log in to reply

Can you explain what "now by keeping factors" mean? – Calvin Lin Staff · 1 year, 1 month ago

Log in to reply

– Lakshya Sinha · 1 year, 1 month ago

That's what I was talking about.Log in to reply

– Nihar Mahajan · 1 year, 1 month ago

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