Quant Boosters  Vikas Saini  Set 4

We know highest power comes by this formula.
Highest power = [n/p]+[n/p^2]+[n/p^3]……..
52 = [n/7] + [n/7^2] + [n/7^3]
The only things strike in our mind to split 52.
1 + 7 + 49 (not equal to 52)
6 + 42 (not equal)
7 + 49 (not equal)
6 + 42 < 52 < 7 + 49
6 + 43 < 52
6 + 46 = 52.
[n/7] = 46
n > 7 * 46.
n > 322.
[n/49] = 6.
n > 49 * 6.
n > 294.
For least possible number n = 322.

Q6) Find the least number which has highest power of 13 as 52.

52 = [n/13] + [n/13^2]
52 = [ n/13] + [n/169]
Let’s think , how 52 can be split.
3 + 39 < 52 < 4 + 52
3 + 49 = 52.
[n/13 ] = 49.
n > 13 * 49
n > 637.
[n/169] = 3
n = 169 * 3
n > 507.
For least value, n = 637.

Q7) Find highest power of 72 in 100!

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!