Find a 5-digit number abcde such that 3 times 1abcde is abcde1.
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Hence the Answer will be ⇒ 4 2 8 5 7
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
Wow.,..what a solution :D
Let: x = abcde
Then,
3(100,000 + x) = 10x + 1
300,000 + 3x = 10x + 1
299,999 = 7x
42857 = x
Final answer: 42857
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^^^^^^
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Let the 5-digit number be N.
3(10^5+N)=10N+1.
This implies 3*10^5-1=299999=7N
Therefore N=42857