# Quant Boosters - Vikas Saini - Set 4

• 72 = 2^3 x 3^2.
Highest power of 2 = [100/2] + [100/2^2] + [100/2^3] + [100/2^4] + [100/2^5] + [100/2^6]
= 50 + 25 + 12 + 6 + 3 + 1
= 97.

But power of 2 is 3 here, so 97/3 = 32.
Highest power of 3 = [100/3] + [100/3^2] + [100/3^3] + [ 100/3^4]
= 33 + 11 + 3 + 1
= 48.

But power of 3 is 2 here, so 48/2 = 24.
24 < 32.
Power of 72 in 100! Is 24.

• Q8) Find the highest power of 24 in 50!

• 24 = 2^3 x 3.
Power of 2 in 50! = [50/2] + [50/2^2] + [50/2^3] +[50/2^4]+[50/2^5]
= 25+12+6+3+1
= 47.
Power of 2 is 3 here.
47 / 3 = 15.

Power of 3 in 50!
[50/3]+[50/3^2]+[3^3]
= 16+5+1
= 21.

15 < 21.
Power of 24 in 50! is 15.

• Q9) Find the total no of divisors of 17!

• Prime factors are 2,3,5,7,11,13,17.
Highest power of 2 = [17/2] + [ 17/2^2] + [ 17/2^3] + [17/2^4]
8+4+2+1 = 17.
Highest power of 3 = [17/3] + [17/3^2]
5+1 = 6.
Highest power of 5 = [17/5] + [17/5^2] = 3.
Highest power of 7 = [17/7] + [17/7^2] = 2.
Highest power of 11 = [17/11] = 1.
Highest power of 13 = [17/13] = 1.
Highest power of 17 = [17/17] = 1.
17! = 2^17 x 3^6 x 5^3 x 7^2 x 11 x 13 x 17
Factors = (17+1)(6+1)(3+1)(2+1)(1+1)(1+1)(1+1)
= 18 x 7 x 4 x 3 x 2 x 2 x 2
= 126 x 96
= 12096.

• Q10) Find the no of zeroes at the end of 100 * 99^2 * 98^3 * 97^4 ... 1^100

• 100 x 99^2 x 98^3 ... 1^100
= (100 x 99 x 98 ... 1) x (99 x 98 x 97... 1) x (98 x 97 x 96 ... 1) ... 1
= 100! x 99! x 98! x 97! ... 1!
= 1! x 2! ... 100!

1! to 4! = 0.
5! to 9! = 1 x 5 = 5.
10! to 14! = 2 x 5 = 10.
15! to 19! = 3 x 5 = 15.
20! to 24! = 4 x 5 = 20.
1! to 24! = 5(0+1+2+3+4) = 50.
25! To 49! = 5(6+7+8+9+10) = 200.
50! to 74! = 5(12+13+14+15+16)=350.
75! to 99! = 5(18+19+20+21+22)=500.
100! = 24.
Total = 50 + 200 + 350 + 500 + 24 = 1124.

• Q11) Find no of zeroes at the end of 250 * 255 * 260 ... 750.

• 5^101 (50 x 51 x 52 …………………150)
= 5^101 (150 !) / (49! )

No of 5’s in 150! = [150 / 5] + [ 150 / 5^2 ] + [ 150 / 5^3 ]
= 30 + 6 + 1 = 37.
Total no of 5’s in numerator = 101 + 37 = 138.

No of 2’s in 150!
= [150 / 2] + [150 /2^2 ] + [ 150 /2^3 ] + [ 150/2^4 ] + [ 150 / 2^5 ] + [ 150 / 2^6 ] + [150 / 2^7]
= 75 + 37 + 18 + 9 + 4 + 2 + 1
= 146.

No of 2’s > No of 5’s
Hence no of zeroes at numerator = no of 5’s = 137.

Number of zeroes at denominator = [ 49 / 5 ] + [ 49 / 5^2 ]
= 9 + 1
= 10.

Total no of zeroes = 137 – 10 = 127.

• Q12) For any value of k, k! has n zeroes at the end and (k+2)! has n+2 zeroes at the end. How many number of possible values of k are there if 0 < k < 200.

• We know no of zeroes increases by 2 only when no is in form of 25n+1.
k+2 = 25n+1.
k+2 = 26,51,76,101,126,151,176,201.
But for 125, no of zeroes increase by 3.
So we need to remove 126.
k+2 = 26,51,76,101,151,176,201
k = 24,49,74,99,149,174,199.
Total possible values = 7.

• Q13) k! has n number of zeroes at the end and (k+1)! has (n+3) zeroes at the end. Find the number of possible values of n if 100 < k < 1000.

• We know for multiple of 5(5^1), no of zeroes increase by 1.
For multiple of 25(5^2), no of zeroes increase by 2.
For multiple of 125(5^3), no of zeroes increase by 3.
It means k+1 = 125 a.
k+1 = 125,250,375,500,625,750,875,1000.
But 625(5^4), no of zeroes increase by 4.
So k+1 = 125,250,375,500,750,875,1000.
Total values of k = 7.

• Q14) Find the highest power of 2 in 128! + 129! + 130! ... 200!

• 128! (1 + 129 + 129 x 128 + 129 x 130 x 131 + 129 x 130 x 131 x 132 ... + 129 x ... 200 )
= 128! ( 130 + 129 x 128 + 129 x 130 x 131 + 129 x 130 x 131 x 132 ...)
= 128! x 2 ( 65 + 129 x 64 + 129 x 65 x 131 + 129 x 65 x 131 x 132 ...)
= 128! x 2 ( 65 + even + odd + odd ....)
= 128! x 2 ( odd + odd + odd ...)

No of 2’s in 128! = [128/2] + [128/2^2] + [128/2^3] + [128/2^4] + [128/2^5] + [128/2^6] + [128/2^7]
= 64 + 32 + 16 + 8 + 4 + 2 + 1
= 127.

But there is one more 2.
So total no of zeroes = 127 + 1 = 128.

• Q15) Find the first non zero digit of 97!

• Concept : If we have to find first non zero digit from right in any number n!
If n = 5 x m + k.
Then 2^m x m! x k!
If n = 25 x m + k (useful for large numbers)
Then first non zero digit = 4^m x m! x k!
Similarly, if n = 125 x m + k. (Again, useful for large numbers)
Then first non zero digit = 8^m x m! x k!

Here, 97 = 25 x 3 + 22.
First non zero digit = 4^3 x 3! x 22!

22! = 5 x 4 + 2.
First non zero digit = 2^4 x 4! x 2!
= 16 x 24 x 2
=8 (first digit from right).

First non zero digit of 97 = 4^3 x 3! x 8
= 4 x 6 x 8
= 2.

Although it can be solved in another way.
97 = 5 x 19 + 2.
Digit = 2^19 x 19! x 2!
19 = 5 x 3 + 4.
Digit = 2^3 x 3! x 4! = 2.
Digit of 97! = 2^19 x 2 x 2
= 8 x 2 x 2
= 2.

One more example

Find the first non zero digit of 377!

377 = 125 x 3 + 2.
First non zero digit = 8^3 x 3! x 2! = 2 x 6 x 2 = 4.

Another way :-
377 = 25 x 15 + 2.
Digit = 4^15 x 15! x 2! .
15! = 3 x 5.
Digit = 2^3 x 3! = 8.
Digit of 377! = 4^15 x 8 x 2 =4.

• Q16) What is the unit digit of 16^48 + 17^48 + 18^48 + 19^48 ?

• We need to take just unit digit of every number.
48 is common as power of each number.
48 mod 4 = 0.
Although remainder is 0 here, we consider it power of 4.
6^(anything) = 6.
7^4 = 1.
8^4 = 6.
To get unit digit of 9, we need to check power is even or odd.
48 is even.
9^even = 1.
Sum of each unit digit number = 6 + 1 + 6 + 1 = 14.
Hence unit digit is 4.

• Q17) Find unit digit of 1^14 + 2^14 + 3^14 … 100^14

63

61

61

58

1

34

61

54