how many prime numbers less that 100 and greater than 3 are of the form:4x+1 , 5y-1

• all the prime numbers between 3 and 100 are:5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73, 79, 83, 89 and 97
  numbers must end with 4 or 9 to follow 5y-1 condition.
  so the numbers are:29,89
  so there are two numbers.
  
• 10 years agoHelpfull: Yes(48) No(11)
• there are 2 prime numbers less than 100 and greater than 3 are of the form of: 4x+1, 5y-1. The numbers are 29 & 89.
  First we take the eqn: 4x+1 So the nos which are greater than 3 & less than 100 are 5,9,13...97. Then we take the prime nos. from there. Then we consider the eqn: 5y-1 and again the nos are: 4,9,14,19,....99. Then we take only the prime nos from there. And lastly we take the common between the two prime nos. set. And we get 29 & 89.
• 10 years agoHelpfull: Yes(29) No(2)
• 14 no.s are 5,13,17,19,29,37,41,57,61,71,73,79,89,97.
• 10 years agoHelpfull: Yes(7) No(7)
• Ans 2
  29,89
  for condition 5y+1--19,29,59,79,89,
  for condition 4x+1---5,13,17,,29,37,41,53,61,88,97.
  For conditions both satisfying 29,89
• 10 years agoHelpfull: Yes(6) No(3)
• ans: 14
  prime nos of the form 4x+1=9
  prime nos of the form 5y-1=5
• 10 years agoHelpfull: Yes(2) No(6)
• only two numbers are there.....29 and 89
• 10 years agoHelpfull: Yes(2) No(0)
• Let the number be N.
  So N = 4x + 1 = 5y - 1
  ⇒x=5y−24
  y = 2 satisfies the equation. So minimum number satisfies both the equations is 9 and general format of the numbers which satisfies the equation = k. LCM (4, 5) + 9 = 20k + 9.
  Now by putting values 1, 2, 3 . . . . for k, we get 29, 49, 69, 89. Of which only 29, 89 are primes.
  
• 9 years agoHelpfull: Yes(2) No(2)
• answer should be 12 and the numbers are 5,13,17,19,29,37,41,53,59,61,79,89
• 10 years agoHelpfull: Yes(1) No(3)
• total no will be 14
  4x+1=9
  5y-1=5
  
• 10 years agoHelpfull: Yes(1) No(3)
• 19 also satisfies both the conditions pls tell me y u all are not considering 19
• 7 years agoHelpfull: Yes(1) No(0)
• Prime Number between 3 to 100 excluding the upper limit is 23.
  Now 5y-1 satisfying numbers are 19,29,59,79,89.
  Now take next equation 4x+1 and equate it with the obtain value of the first equation because both will satisfy the given number.
  4x+1 = 19, 4x+1 = 29, 4x+1 = 59, 4x+1 = 79, 4x+1 = 89 and after equating we will get only 2 number which satisfy it.
  Hence the number is 2
  i.e 29,89 Ans
• 5 years agoHelpfull: Yes(1) No(0)
• 29
  4(8)+1=5(6)-1
• 10 years agoHelpfull: Yes(0) No(5)
• total no will be 14
  4x+1=9
  5y-1=5
• 10 years agoHelpfull: Yes(0) No(6)
• 29,19,79,59,89 so 5 nos
• 10 years agoHelpfull: Yes(0) No(4)
• 5X-1 put x= 1 to 100
  so we ill get the values ending with 4, 9
  4X+1 put x =1 to 100
  we will get 5,9,13,17,21,25,29,33,.........97
  from both of these conditions 89 is common .89 satisfies both conditions
  so ans is 89
• 8 years agoHelpfull: Yes(0) No(2)
• ,ll,mkl,l;
• 5 years agoHelpfull: Yes(0) No(1)