# Quant Boosters - Vikas Saini - Set 4

• 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

• 14 mod 4 = 2.
We need to find unit digit number just from 1 to 9,then multiply by 10.
Unit digit of 1,5,6 is 1,5,6 respectively.
Unit digit of 2,3,7,8 is 4,9,9,4 respectively.
Unit digit of 4 and 9 is 6 and 1.
Add all = 1+5+6+4+9+9+4+6+1 = 45.
We will multiply it by 10 because unit digit will be as same as 1 to 9 of 11 to 19,21 to 29 and so on.
Now 45 x 10 = 450.
Hence unit digit 0.

• Q18) Find unit digit of ((1997)^41)^401

• Unit digit is 7.We have to use cycle of 4.
41^401 mod 4 = 1.
7^1 = 7.
Unit digit is 7.

• Q19) Find the last two digits of 147^912

• Concept - To get unit digit number we see only unit digit whereas to get last two digits, we see only last two digits.
If any number’s unit digit is 1, then to get last two digits we apply this formula
(xyz1)^abc = (c x z)/1 (last two digit)
Then unit digit is nothing but 1.
Ten’s digit = z x c.

Examples -
Find the last two digits of 21^3.
Unit digit is 1.
Ten’s digit = 2 x 3 = 6.
Hence last two digits of 21^3 is 61.
(21^3 = 9261).

Find the last two digit of 91^2.
Unit digit is 1.
Ten’s digit = 9 x 2 = 18 = 8.
Hence last two digits of 91^2 = 81.
(91^2 = 8281)

Shortcut approaches we will apply here :-

1. Try to get unit digit 1, so that we could apply above written formula.
Though it’s possible in case of odd integers only.
2. We need to find remainder of power after dividing by 40 only.
3. In case of even number, we shall try to make power of 2.
2^10 = 24 (last two digits)
2^20k = 76(last two digits)

In our given question, We need to focus on 47 from 147.
We know for 7, cycle of 4 works.
7^4 = 1(unit digit)
Hence 47^4 = 81(last two digits).
912 mod 40 = 32.
32 = 4 x 8.
Therefore 47^912 = 81^8.
We have unit digit 1,we can apply formula here.
Unit digit = 1.
Ten’s digit = 8 x 8 = 64 = 4.
Hence last two digits = 41.

• Q20) Find the last two digits of 296^144

42

44

61

61

61

64

63

71