Finding the last 3 digits

May 2015
78
7
Australia
Question:
The first digit of a six-digit number is 1. This digit 1 is now moved from the first digit position to the end, so it becomes the last digit. The new six-digit numbers is now 3 times larger than the original number. What are the last three digits of the original number?

I attached the book's answer solutions to the post.

I understand the 2nd answer alternative.

But for the first answer alternative, I got confused when they multiplied the left side by 10 and the right side by 10^5
 

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May 2016
1,310
551
USA
$n = original\ number.$

$Let\ y = 10000a + 1000b + 100c + 10d + e.$

$n = 100000 + 10000a +1000b + 100c + 10d + e = 100000 + y \implies 3n = 300000 + 3y.$

$BUT\ 3n = 100000a + 10000b + 1000c + 100d + 10e + 1 = 10y + 1.$

$\therefore 10y + 1 = 300000 + 3y \implies 10y - 3y = 300000 - 1 \implies$

$7y = 299999 \implies y = \dfrac{299999}{7} = 42857 \implies$

$n = 100000 + 42857 = 142857.$ Clear now? Horrible notation used in the explanation of the first alternative.
 
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Denis

Math Team
Oct 2011
14,592
1,026
Ottawa Ontario, Canada
Couple of other cases that are somewhat similar:

285714 : 857142 (1st digit moved to end is 2)

410256 : 102564 (4 is digit moved: 102564 * 4 = 410256)

Interesting: 142857 * 5 = 714285
 
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