A teachers was trying to form the groups of students in such a way that every group has equal number of student in such a way that every group has equal number of students and that number be a prime number.She tried for first 5 prime numbers but on each occasion exactly one student was left behind if the numbers of students is in 4 digits then how many different can the total no of students take?
A.0
B.2
C.3
D.4
Answers
Answer:
4
Step-by-step explanation:
first 5 prime numbers
2
3
5
7
11
LCM of these
2*3*5*7*11= 2310
on each occasion exactly one student was left behind
so number of possible students
2310+1=2311
2310*2 +1=4621
2310*3+1= 6931
2310*4+1=9241
2310*5+1 will be 5 digits number
so 4 digits numbers can be 4.
Answer:
4
Step-by-step explanation:
First 5 prime numbers = 2, 3, 5, 7 and 11
We have to find find lcm (2, 3, 5, 7, 11)
LCM = 2 × 3 × 5 × 7 × 11 = 2310
It is given that on each occasion exactly one student was left
Now for getting remainder as 1 in each case we have to add 1 in it
So, the number of possible students
2310 × 1 + 1 = 2311
2310 × 2 + 1 = 4621
2310 × 3 + 1 = 6931
2310 × 4 + 1 = 9241
2310 × 5 + 1 = 11551 this is 5 digit number.
It is clear that the 4 digits numbers are only 4.