The largest number amongst the following that will perfectly divide (101^100)-1 is..

1) 100
2) 10,000
3) 100,000
4) 100^100

• 2. 10, 000
  
  101 x 101 - 1 = 10201 - 1 = 10200 Divisible by 100
  
  101 x 101 x 101 - 1 = 1030301 - 1 = 1030300
  
  We can observe that 101 ^ 9 - 1 will give 2 zero's at the end & are safely divisble by 100.
  So 101 ^ 10 - 1 to 101 ^ 99 -1 is safely divisble by 1000
  So 101 ^ 100 - 1 will be divisible by 10000
• 12 years agoHelpfull: Yes(6) No(1)
• (101^2)-1 = 10200
  (10200^2)-1 = 104050400 . . . . .
  last two digits are 0
  so ans might be 100
• 12 years agoHelpfull: Yes(1) No(5)
• Let me first appreciate the beauty of the question, it's a very good question.
  we have 101^100-1
  Let's keep 1 aside.
  101^100 can be written as (100+1)^100
  After binomial expansion:
  101^100 =100C0 *(100)^100*1^0 +100C1 *( 100)^99*1^1 + ...... +100C99*(100)^1*1^99+100C100*(100)^0*1^100
  for any number to be divisible by any other number its ''sum representation " must be divisible by the divisor.
  so in above expression, we can easily see each term is divisible by 10000 except for the last term which is 1
  so the expression 100^101-1 is divisible by 10000 and is the maximum factor.
  
  
  Note: In the case of 100^100 which is a tempting option , we can easily see, it appears in Binomial Expansion but each term of the binomial expansion is not divisible by it.
  for example the second term itself i.e. 100C1*100^99*1^1=100C1*100^99=100^100 (divisible)
  but third term 100C2*100^98*1^2={100*99/2!}*100^98=100^99*{99/2} ---- clearly non-divisible.
  Students are advised not to be tempted by the options and do check options with utmost sincerity.
  
  
  THIS QUESTIONS CAN BE SOLVED IN SECONDS WITHOUT BINOMIAL EXPANSION WITH THE GIVEN OPTIONS
  Applying remainder theorem for 100^100 {since maximum divisor has been asked so it's logical to check options in descending order}
  101^100-1 { 101}^100
  ---------------- = ---- minus 1/(100^100) NOT DIVISIBLE
  100^100 { 100}^100
  
  Apply remainder theorem for 10000 { second largest no in option}
  
  101^100-1 10201^99-1 201^99-1 20301^98-1 301^98 -1
  --------------= -------------- = ------------- = ------------------ = ----------
  10000 10000 10000 10000 10000
  
  
  and so on until we are left with
  901 raised to some power divided by 10000
  then the next step will make it
  91001 raised to some power
  and then 1002001 raised to some power divided by 10000
  and finally we will reach 9001raised to some power divided by 10000
  i.e 9010001 raised to some power divided by 10000
  which gives 1 divided by 10000
  but 1 is reduced from 101^100 to encounter for complete expression.
  
• 8 years agoHelpfull: Yes(0) No(2)