Last two digits of 87^474?

• (87^4)^118*(87)^2
  =(61)^118*(7569)
  =481*7569
  89
  last two digits are 89
• 8 years agoHelpfull: Yes(9) No(2)
• (87^4)^118 * 87^2
  =(87^2 * 87^2)^118 * 87^2
  now 87^2=7569 we will take only last two digit number i.e. 69
  =(69 * 69)^118 *87^2
  69 *69 gives unit digit only=01 and 87^2 gives unit digit 9
  01 * 9= 09
  so, last two digit= 09
  
• 8 years agoHelpfull: Yes(3) No(5)
• From the cyclicity of 7,It repeats after 4 intervals. now 474/4 makes remainder 5.
  so last digit in thw extreme rigit is 5.
  again from the cyclicity of 87 ,it repeats after 2 intervals. but 474 /2 makes the remainder 0.
  so we takes as 87^2---- last digit 9.
  ultimately the answerwill be 95
• 8 years agoHelpfull: Yes(1) No(7)
• http://www.slideshare.net/nickyswetha/number-systems-theory-for-cat-2009-quant please have a luk at this link for clear solution
• 8 years agoHelpfull: Yes(1) No(0)