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

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

Asked In TCS harti (9 years ago)

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shashank shekhar   (9 years ago)

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.

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Sayantonee Maitra  (9 years ago)

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.

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Sudhanshu Ranjan  (9 years ago)

14 no.s are 5,13,17,19,29,37,41,57,61,71,73,79,89,97.

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Narasapu Hari prasad  (9 years ago)

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

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soumya  (9 years ago)

ans: 14
prime nos of the form 4x+1=9
prime nos of the form 5y-1=5

Is this solution Helpfull? Yes (2) | No (6)  

bujji  (9 years ago)

only two numbers are there.....29 and 89

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sowmya  (8 years ago)

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.

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PRANAVI MANCHIRAJU  (9 years ago)

answer should be 12 and the numbers are 5,13,17,19,29,37,41,53,59,61,79,89

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Chirag Ranjan Lahiri  (9 years ago)

total no will be 14
4x+1=9
5y-1=5

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Abhirami  (6 years ago)

19 also satisfies both the conditions pls tell me y u all are not considering 19

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veda  (4 years ago)

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

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anushakola  (9 years ago)

29
4(8)+1=5(6)-1

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Aastha  (9 years ago)

total no will be 14
4x+1=9
5y-1=5

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prashansa  (9 years ago)

29,19,79,59,89 so 5 nos

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ramadevi  (7 years ago)

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

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pawan verma  (4 years ago)

,ll,mkl,l;

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