Find last two digits of the following expression (201*202*203*204*246*247*248*249)^2

• (01*02*03*04*46*47*48*49)^2
  (01*04*09*16*16*09*04*01) on sq.
  (01*04*09*16)^2
  (576)^2
  Last 2 digit is 76
• 8 years agoHelpfull: Yes(3) No(0)
• To find th last two digit the number must be divided by 100 and the remainder will be the ans .
  now (201*202*203*204*....248*249)divided by 100
  remainder=(1*2*3*....48*49)=49!
  from online calculator u can easily calculate the last two digit of 491 i.e 24
  
  now the no is in whole square form so it will be (24)^2=576
  so the last two digit is 76
• 11 years agoHelpfull: Yes(2) No(8)
• Answer will be 00
• 11 years agoHelpfull: Yes(2) No(2)
• (201*202*203*204*246*247*248*249)=201*202*203*204*246*247*248*249 mod 100
  = 1*2*3*4*46*47*48*49 mod 100
  = 98*192*141*46 mod 100
  = -2*-8*41*46 mod 100
  =16*41*46 mod 100
  =92*8*41 mod 100
  = 92*328 mod 100
  = -8*28 mod 100
  =-224 mod 100
  =76 mod 100
  
  Now if (201*202*203*204*246*247*248*249) is 76 mod 100
  (201*202*203*204*246*247*248*249)^2 = 76^2 mod 100
  =5776 mod 100
  = 76
• 8 years agoHelpfull: Yes(0) No(1)