Quant Boosters - Vikas Saini - Set 4

• 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 ?

• Number\BallsRedBlackGreenWhite
11234
25678
39101112
413141516
517181920
621222324
725262728
8292--
Total120100105112

Total = 120 + 100 + 105 + 112 = 437.

437th ball is the 100th black ball.

• Q30) If A, B, C, D, E represent distinct digit from 1 to 5.
ABE = E x BE.
CDA = CA x AA.
BCA + CCD = EEE.
Find the value of AB x CE.

43

63

31

61

62

61

61