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Logic Level 3

Find the 6-digit number that increases 6 times when its last 3 digits are carried to the beginning of the number without their order being changed.

The answer is 142857.

1 solution

Kazem Sepehrinia
Aug 13, 2015

Let that number be $\overline{ABCDEF}$ , we have $\overline{DEFABC}=6\overline{ABCDEF} \\ 1000 \overline{DEF}+\overline{ABC}=6000 \overline{ABC}+6 \overline{DEF} \\ 994 \overline{DEF}= 5999 \overline{ABC} \\ \frac{\overline{ABC}}{\overline{DEF}}=\frac{142}{857}$ Thus, $\overline{ABCDEF}=142857$ .

Moderator note:

That's wonderful!

Dude, Thats a nice question, i appreciate that.. But, please edit the question so that its not confusing.. Increases by 6 times mean that the value becomes 7 times the original value.. I hope it makes sense..

Harshavardhan Resdy - 5 years, 6 months ago

Note than 1/7=0.142857

Hasmik Garyaka - 2 years, 6 months ago

Why did you multiply $\overline{DEF}$ and $\overline{ABC}$ by 1000?

Vishal Yadav - 5 years, 7 months ago

Say DEFABC = 123456 then we can write DEFABC = 123000 + 456 =(123)(1000)+456 So we can write DEFABC = (DEF)(1000)+ABC or 1000DEF + ABC

Mahabubul Islam - 5 years, 6 months ago

Thanks but Beware on Brilliant. Asking a question can give you 5 downvotes.

Vishal Yadav - 5 years, 6 months ago

Come on! You can do it!

The answer is 142857.

1 solution

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