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Home » Mathematics » Middle School Classroom Math (Ages 9-13)
 
A five-digit number
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small_unicorn
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#1
A five-digit number
There is an interesting five-digit number. With a 1 after it, it is three times as lardge as with a 1 before it. What is it ?

from book "The Moscow puzzles" by Boris A. Kordemsky

Please hide Your solution . Write inside Hide ... /Hide

Wink

PostPosted: Tue Nov 03, 2009 4:05 pm  Back to top 
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gaussintraining
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#2
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42857 Mr. Green


PostPosted: Tue Nov 03, 2009 4:11 pm  Back to top 
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small_unicorn
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#3
gaussintraining wrote:
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42857 Mr. Green

Solution ?
Juggle

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gaussintraining
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#4
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142857 or the repeating part of \frac{1}{7}=\overline{.142857}, has many interesting properties relating to the fact that each of its first 6 multiples has the same digits as the others, just in a different order.

142857

285714

428571

571428

714285

857142

When the question stated that it was a five digit number, which when adding a digit to the front or back of it resulting in 2 numbers, with the property that one was 3 times the other (and of course noting that a 5 digit number with another digit makes a 6 digit number), I thought of this sequence of numbers, since it is often the solution to puzzles of that kind with 6 digit numbers and multiples that can be rearranged. A quick check confirmed that 142857\times{3}=428571. So the original 5 digit number was \boxed{42857}


PostPosted: Tue Nov 03, 2009 4:51 pm  Back to top 
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steve123456
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#5
I found a different way.Here it is.

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Call the number abcde, with the variables the digits. Therefore, with what is given, you can conclude that abcde1=3(1abcde). You can write abcde1 and 1abcde in expanded form. This is what abcde1 would be like in expanded form: 100,000a + 10,000b + 1,000c + 100d +10e + 1. 1abcde would be: 100,000 + 10,000a + 1,000b + 100c + 10d + e.
By substitution you can form the conclusion that 100,000a + 10,000b + 1,000c + 100d + 10e + 1=3(100,000 + 10,000a + 1,000b + 100c + 10d + e.) Distributing gives 100,000a + 10,000b + 1,000c + 100d + 10e + 1=300,000 + 30,000a + 3,000b + 300c + 30d + 3e. Subtracting from both sides forms 70,000a + 7,000b + 700c + 70d + 7e= 299,999. Dividing by 7 yields 10,000a + 1,000b + 100c + 10d + e=42857. Taking this and putting it together makes abcde. Therefore, abcde=42857


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PostPosted: Thu Nov 05, 2009 5:40 pm  Back to top 
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MongolJohn
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#6
My solution started similarly to Steve's:

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Let x = the 5-digit number

10x + 1 = 3(100000 + x)
10x + 1 = 300000 + 3x
7x = 29999
x = 42857



I just found this forum, and I'm liking it a lot!

PostPosted: Fri Nov 13, 2009 1:40 pm  Back to top 
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