# Quant Boosters - Vikas Saini - Set 4

• 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

• We take only 96.
As we can’t get 1 in any power of 6.
So 144 mod 40 = 24.
Therefore 296^144
= 96^24
= 16^12
= ((2)^4)^12
= 2^48
= 2^(20k+8 )
= 2^20k x 2^8
= 76 x 56
= 56.
Hence last two digits are 56.

• Q21) Find last two digits of 488^488

• We take 88 from 488.
488 mod 40 = 8.
= 88^8
= (2^3 x 11 )^8
= 2^24 x 11^8
= 2^(20K+4) x 11^8
= 2^20k x 2^4 x 11^8
= 76 x 16 x 81
= 16 x 81
= 96
Hence last two digits 96.

• Q22) Find the last two digits of 297^455

• We take 97 from 297.
455 mod 40 = 15.
Therefore 297^455
= 97^15
= (97)^4k + 3
= (97^4)^3 x 97^3
=81^3 x 97^2 x 97
= 41 x 09 x 97
= 69 x 97
= 93.
Hence 93.

• Q23) In first 252 natural numbers, how many times we need to write number 4?

• Unit place = 10+10+5=25.
Ten’s place = 10+10+10 = 30.
Total times = 25 + 30 = 55.

• Q24) In first 252 natural numbers, how many numbers contain digit 4?

• Unit digit = 10 + 10 + 5 = 25.
Ten’s digit = 10 + 10 + 10 = 30.
Unit & Ten’s digit both = 3 (44,144,244)
Total numbers = 25 + 30 - 3 = 52.

• Q25) In book has 100 pages, how many digits have been used to number the pages ?

• 1- 9 = 9 digits
10 – 99 = 2 x 90 = 180 digits.
100 = 3 digits.
Total = 9 + 180 + 3 = 192.

• Q26) There are 3 consecutive natural numbers such that square of the second is 19 more than 9 times of sum of first & third number. Find second number.

• Let 3 consecutive numbers are n-1, n and n+1.
=> n^2 – 19 = 9(n -1+n+1)
=> n^2 – 19 = 9 x 2n
=> n^2 – 18n + 19 = 0.
=> (n-19) (n+1) = 0.
=> n = 19, -1.
Second number is 19.

• Q27) Let N = 999999 ... 36 times. How many 9’s are there in N^2 ?

• 99^2 = 9801 (one 9)
999^2 = 998001 (two 9)
9999^2 = 99980001 (three 9)
Therefore (99999……….36 time)^2 should have 35 nines.

• Q28) How many positive numbers are there which is equal to 12 times sum of their digits ?

• For 2 digit number
Let the number is 10a+b.
10a + b = 12(a+b)
= > 10a + b = 12a + 12b
= > -2a = 11b
None two digit number exist here.

For 3 digit numbers
Let the number is 100a+10b+c.
100a + 10b + c = 12(a+b+c)
= > 100a + 10b + c = 12a + 12b + 12c
= > 88a – 2b = 11c.
b = 0.
= > 88 a = 11 c.
= > 8a = c.
a = 1, c = 8.
Number 108.

For 4 digit number
1000a+100b+10c+d
1000a+100b+10c+d = 12(a+b+c+d)
= > 1000 a + 100 b + 10 c + d = 12 a + 12 b + 12 c + 12 d
= > 988 a + 88b = 2c+11d
Not possible.

Hence only positive number is 108.

• Q29) If we arrange 4 different color balls like
1 red ball, 2 black balls, 3 green balls, 4 white balls
5 red balls, 6 black balls, 7 green balls, 8 white balls etc...
How many balls would be required to reach the 100th black ball ?

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