Not so difficult.

Find a 5-digit number abcde such that 3 times 1abcde is abcde1.


The answer is 42857.

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5 solutions

George Darroch
Nov 1, 2014

Let the 5-digit number be N.

3(10^5+N)=10N+1.

This implies 3*10^5-1=299999=7N

Therefore N=42857

Nafees Zakir
Oct 13, 2014
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for a in range(1,10):
    for b in range(1,10):
        for c in range(1,10):
            for d in range(1,10):
                for e in range(1,10):
                    if(3*(100000+a*10000+b*1000+c*100+d*10+e)==(100000*a+b*10000+c*1000+d*100+e*10+1)):
                        print "%d%d%d%d%d"%(a,b,c,d,e)

42857

Hence the Answer will be 42857 \Rightarrow\boxed{42857}

1a b c d e × 3 = a b c d e 1

e × 3 = 1 in unit place so e = 7

3 × d + 2 (c/f) = e = 7 in unit place so 3 × d = 5 then d = 5

3 × c + 1 (c/f) = d = 5 so 3 × c = 4 in unit place then c = 8

3 × b + 2 (c/f) = c = 8 so 3 × b = 6 in unit place then b = 2

3 × a = b = 2 in unit place then a = 4

so a b c d e = 4 2 8 5 7

Sunil Pradhan - 6 years, 7 months ago

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i too did it in same way....

sadaram bhuvanesh - 6 years, 6 months ago

Wow.,..what a solution :D

Krishna Ar - 6 years, 7 months ago
Christian Daang
Nov 4, 2014

Let: x = abcde

Then,

3(100,000 + x) = 10x + 1

300,000 + 3x = 10x + 1

299,999 = 7x

42857 = x

Final answer: 42857

Amos Tan
Jun 12, 2015

1abcde * 3 = abcde1>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> (100000+ 10000a +1000b+ 100c+ 10d +e) *3= 100000a +10000b +1000c + 100d +10e +1 >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> 300000 + 3(10000a +1000b+ 100c+ 10d +e) =10( 10000a +1000b+ 100c+ 10d+ e ) >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> 299999 = 7( 10000a +1000b+ 100c+ 10d +e )>>>>>>>>>>>>>>>>>>>>>>>>>> 42857 = 10000a +1000b+ 100c+ 10d +e ^^^^^^a=4, b=2, c=8, d=5, e=7^^^^^^

Raj Thaker
Dec 9, 2014

That's easy $MONEY$

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