Find the last non-zero digit of 30^2345

• 3^1=3;
  3^2=9;
  3^3=27;
  3^4=81;
  3^5=243; so it is a power cycle of 4, (9,7,1,3 will be repeating ...) so
  2345/4 gives you 1; so 3^1= 3 so 3 is the remainder..
• 9 years agoHelpfull: Yes(13) No(1)
• 30^2345=(3*10)^2345=3^2345*10^2345
  so 30^2345 has 2345 zeroes,the last non-zero digit of 30^2345 will be same as unit digit of 3^2345
  3^2345=3^(4*586+1)=> 3^1=3
  last non-zero digit of 30^2345= 3
• 10 years agoHelpfull: Yes(11) No(1)
• 3
  eg: 30^5
  30x30=900
  900x30=27000
  27000x30=810000
  810000x30=24300000
  
  similarly at 2345th time , it wud b 3.. coz 2345 s a multiple of 5
  
• 10 years agoHelpfull: Yes(5) No(1)
• Ans : 3
  30^2345 = 3^2345 * 10^2345
  3^1=3 3^2=9 3^3=27 3^4=81 then it will repeat!
  So 2345%4 would give a remainder of 1 and hence the remainder is 3!
• 8 years agoHelpfull: Yes(0) No(0)