Sir how to solve these problems

If 100! divisible by 3^n
then find the maximum
value of n.

(a) 48 (b) 44
(c) 40 (d) 33
24. If 122! is divisible by 6^n
then find the maxi-
mum value of n.

(a) 58 (b) 62
(c) 40 (d) 48

Q: If 100! divisible by 3^n then find the maximum value of n.

(a) 48 (b) 44
(c) 40 (d) 33

Solution: 100 is the multiple of 2 and so the occurrence of two will be high. so if occurrence of 2 > 3 so will will find the occurrence of 3.

we are doing this because we cannot brake 3 in factors if we could then we can take the higher factor of 3

3 100
3 33
3 11
3 3
1

now if you sum 33+11+3+1 the you will get 48 as the power of 3. So we can say that the maximum power of 3^n is 48

Q: If 122! is divisible by 6^n then find the maximum value of n.

(a) 58 (b) 62
(c) 40 (d) 48

Solution: Now see we can brake 6 into 3*2 so 3 is a higher factor of 6 now find the occurrence of 2 in 122

2 122
2 61
2 30
2 15
2 7
2 3
1

Sum these 61+30+15+7+3+1=117 now divide 117 by 4 you will get 58 as a answer.

Please follow Abhinay Sir videos on youtube for better understanding.

x^2-5y^2=1232 how many pairs are possible for (x,y)

If x^2-5y^2=1232 how many pairs are pissible for (x,y)

sir please provide the solution also