8.6 Number System -- Practice MCQs

(b) 1275

Sum = n(n+1)/2 = 50 x 51 / 2 = 1275

(c) 97

87 = 3 x 29  (not prime)
91 = 7 x 13  (not prime)
93 = 3 x 31  (not prime)
97 --> test primes up to sqrt(97) ~ 9.8: {2, 3, 5, 7}
      not divisible by any --> 97 is prime

(b) 3

Cycle of 7: {7, 9, 3, 1}, length = 4
95 mod 4 = 3
3rd position in cycle = 3

Verification: 7^1=7, 7^2=49(unit 9), 7^3=343(unit 3)

(c) 19

floor(80/5) + floor(80/25) + floor(80/125)
= 16 + 3 + 0
= 19

(c) Both 3 and 5

Divisibility by 5: Last digit = 5 --> divisible by 5  ✓
Divisibility by 3: Sum = 7+2+4+8+1+5 = 27, 27/3 = 9 --> divisible by 3  ✓

(b) 12

96 = 2^5 x 3^1
Factors = (5+1)(1+1) = 6 x 2 = 12

(c) 4000

In 3,04,759 the digit 4 is in the thousands place.
Place value = 4 x 1000 = 4000

(c) (16, 21)

(12, 18) --> HCF = 6  (not co-prime)
(15, 25) --> HCF = 5  (not co-prime)
(16, 21) --> 16 = 2^4, 21 = 3 x 7, no common factor --> HCF = 1  ✓
(14, 35) --> HCF = 7  (not co-prime)

(b) 104

Smallest 3-digit number = 100
100 / 8 = 12.5
ceil(12.5) = 13
8 x 13 = 104

(a) 1

345 = 8 x 43 + 1
Remainder = 1

(b) Whole number but not a natural number

Natural numbers: {1, 2, 3, ...}
Whole numbers: {0, 1, 2, 3, ...}
0 is whole but not natural.

(a) 56

28 = 2^2 x 7^1
Sum = [(2^3 - 1)/(2-1)] x [(7^2 - 1)/(7-1)]
    = [7/1] x [48/6]
    = 7 x 8 = 56

Note: 28 is a perfect number (sum of proper divisors = 28),
so sum of ALL divisors = 28 + 28 = 56.

(b) 25

Between 11 and 61 (exclusive): 12 to 60
First even = 12, Last even = 60
Count = (60 - 12)/2 + 1 = 48/2 + 1 = 25

(b) 12

LCM(4, 6) = 12
If a number is divisible by both 4 and 6, it is divisible by their LCM = 12.
Not necessarily by 24: for example, 12 is divisible by 4 and 6 but not by 24.

(c) 1826

1296 = 36^2  ✓
1764 = 42^2  ✓
1826: sqrt(1826) ~ 42.7 (not an integer) --> NOT a perfect square
2025 = 45^2  ✓

Also: 1826 = 2 x 11 x 83 (all exponents are 1, not even) --> not a perfect square.

(c) 30

720 = 2^4 x 3^2 x 5^1
Factors = (4+1)(2+1)(1+1) = 5 x 3 x 2 = 30

(c) 4

2024 mod 4 = 0, so use the 4th (last) element in each cycle.

2^2024: cycle {2,4,8,6}, position 4 --> unit digit 6
3^2024: cycle {3,9,7,1}, position 4 --> unit digit 1
7^2024: cycle {7,9,3,1}, position 4 --> unit digit 1
8^2024: cycle {8,4,2,6}, position 4 --> unit digit 6

Sum of unit digits = 6 + 1 + 1 + 6 = 14
Unit digit of 14 = 4

(d) 9

9 mod 10 = -1 mod 10
(-1)^99 = -1
-1 mod 10 = 9

Alternatively: cycle of 9 is {9, 1}, length 2.
99 mod 2 = 1 --> position 1 = 9.

(a) 85

By inclusion-exclusion:
Divisible by 3:  floor(200/3) = 66
Divisible by 7:  floor(200/7) = 28
Divisible by 21: floor(200/21) = 9

Answer = 66 + 28 - 9 = 85

(b) 47

floor(50/2) + floor(50/4) + floor(50/8) + floor(50/16) + floor(50/32)
= 25 + 12 + 6 + 3 + 1
= 47

(c) 726352

For 726352:
Positions from left: 7(1) 2(2) 6(3) 3(4) 5(5) 2(6)
Odd-position sum:  7 + 6 + 5 = 18
Even-position sum: 2 + 3 + 2 = 7
Difference = 18 - 7 = 11 --> divisible by 11  ✓

Check (a) 123456: (1+3+5)-(2+4+6) = 9-12 = -3 --> No
Check (b) 908712: (9+8+1)-(0+7+2) = 18-9 = 9 --> No
Check (d) 524863: (5+4+6)-(2+8+3) = 15-13 = 2 --> No

(a) 12

360 = 2^3 x 3^2 x 5^1
Number of factors = (3+1)(2+1)(1+1) = 24
360 is NOT a perfect square (exponent 3 is odd).
Number of ways = 24/2 = 12

(c) 20

n(n + 1) = 420
Try n = 20: 20 x 21 = 420  ✓

(b) 40

100 = 2^2 x 5^2
phi(100) = 100 x (1 - 1/2) x (1 - 1/5)
         = 100 x 1/2 x 4/5
         = 40

(b) 1

By Fermat's Little Theorem (17 is prime):
2^16 mod 17 = 1

256 = 16 x 16
2^256 = (2^16)^16 = 1^16 = 1

Remainder = 1

(b) 8

540 = 2^2 x 3^3 x 5^1

Odd factors (ignore the power of 2):
= (3+1)(1+1) = 4 x 2 = 8

(a) 9944

9999 / 88 = 113.625
floor(113.625) = 113
88 x 113 = 9944

(a) 3

For n >= 5, n! contains both 2 and 5 as factors, so n! mod 10 = 0.
Only need: 1! + 2! + 3! + 4! = 1 + 2 + 6 + 24 = 33

33 mod 10 = 3

(b) 3

Sum of known digits = 8 + 9 + 7 + 1 + 5 + 3 = 33
Total sum = 33 + *

For divisibility by 9: (33 + *) must be a multiple of 9.
Next multiple of 9 after 33 is 36.
* = 36 - 33 = 3

(d) 12

For a factor to be a perfect square, each exponent must be even.

For 2^4: even exponents from 0 to 4 --> {0, 2, 4} --> 3 choices
For 3^3: even exponents from 0 to 3 --> {0, 2}    --> 2 choices
For 5^2: even exponents from 0 to 2 --> {0, 2}    --> 2 choices

Total = 3 x 2 x 2 = 12

(c) 3

Step 1: 7 mod 5 = 2, so find 2^(7^7) mod 5.

Step 2: Cycle of 2 mod 5: {2, 4, 3, 1}, length = 4.
        Need 7^7 mod 4.

Step 3: 7 mod 4 = 3, so 7^7 mod 4 = 3^7 mod 4.
        3 mod 4 = -1, so (-1)^7 = -1 = 3 mod 4.

Step 4: 2^(7^7) mod 5 = 2^3 mod 5 = 8 mod 5 = 3.

Answer: 3

(a) 100

3-digit multiples of 6:
  First = 102 (6 x 17), Last = 996 (6 x 166)
  Count = 166 - 17 + 1 = 150

3-digit multiples of 18 (divisible by both 6 and 9):
  First = 108 (18 x 6), Last = 990 (18 x 55)
  Count = 55 - 6 + 1 = 50

Divisible by 6 but not 9 = 150 - 50 = 100

(a) 01

phi(100) = 40
GCD(3, 100) = 1
By Euler's theorem: 3^40 mod 100 = 1

200 = 40 x 5
3^200 = (3^40)^5 mod 100 = 1^5 = 1

Last two digits = 01

(b) 4

10800 = 2^4 x 3^3 x 5^2

For a factor to be a perfect cube, each exponent must be a multiple of 3.

For 2^4: multiples of 3 in {0,1,2,3,4} --> {0, 3} --> 2 choices
For 3^3: multiples of 3 in {0,1,2,3}   --> {0, 3} --> 2 choices
For 5^2: multiples of 3 in {0,1,2}     --> {0}    --> 1 choice

Total = 2 x 2 x 1 = 4

The 4 perfect cube factors: 1, 8, 27, 216

(c) 4

Step 1: 32 mod 7 = 4. Need 4^(32^32) mod 7.

Step 2: By Fermat (7 is prime): 4^6 mod 7 = 1.
        Need 32^32 mod 6.

Step 3: 32 mod 6 = 2. Need 2^32 mod 6.
        Powers of 2 mod 6: 2^1=2, 2^2=4, 2^3=2, 2^4=4, ...
        Even powers give 4. Since 32 is even: 2^32 mod 6 = 4.

Step 4: 4^4 mod 7 = 256 mod 7 = 4   (since 7 x 36 = 252, 256-252 = 4)

Answer: 4

(d) 12

30 = 2 x 3 x 5

A factor of N that is a multiple of 30 must have at least 2^1, 3^1, and 5^1.

Method: Count factors of N/30.
N/30 = (2^3 x 3^2 x 5^2) / (2 x 3 x 5) = 2^2 x 3 x 5

Number of factors of 2^2 x 3 x 5 = (2+1)(1+1)(1+1) = 3 x 2 x 2 = 12

Answer: 12

(a) 5

For n >= 7, n! contains 7 as a factor, so n! mod 7 = 0.
We only need: 1! + 2! + 3! + 4! + 5! + 6!
= 1 + 2 + 6 + 24 + 120 + 720
= 873

873 / 7 = 124 remainder 5  (7 x 124 = 868, 873 - 868 = 5)

Answer: 5

(c) 24

For a factor to be a perfect square, each exponent must be even.

2^6: even exponents from 0 to 6 = {0, 2, 4, 6} --> 4 choices
3^4: even exponents from 0 to 4 = {0, 2, 4}    --> 3 choices
5^2: even exponents from 0 to 2 = {0, 2}        --> 2 choices

Total = 4 x 3 x 2 = 24

(a) 24

Co-primes of 12 less than 12: {1, 5, 7, 11}

Sum = 1 + 5 + 7 + 11 = 24

Formula check: N x phi(N) / 2 = 12 x 4 / 2 = 24  ✓

(a) 400

By inclusion-exclusion:
Divisible by 2: 500
Divisible by 5: 200
Divisible by 10 (both): 100

Divisible by 2 or 5 = 500 + 200 - 100 = 600
Neither = 1000 - 600 = 400

(b) 3

By Fermat's Little Theorem: 2^12 mod 13 = 1

1000 = 12 x 83 + 4
2^1000 = (2^12)^83 x 2^4 = 1 x 16

16 mod 13 = 3

(d) 7, 11, and 13

abcabc = abc x 1000 + abc = abc x 1001
1001 = 7 x 11 x 13

So every number of this form is divisible by 7, 11, and 13.

(a) 5

2^10 = 1024 ~ 10^3
2^30 = (2^10)^3 ~ (10^3)^3 = 10^9
2^29 = 2^30 / 2 ~ 10^9 / 2 = 5 x 10^8

Exact: 2^29 = 536,870,912 --> leading digit = 5  ✓

(c) 8

2160 = 2^4 x 3^3 x 5^1

Total prime factors with repetition = 4 + 3 + 1 = 8
(Distinct prime factors = 3, but with repetition = 8)

(a) 7

146 mod 13: 13 x 11 = 143, remainder = 3
150 mod 13: 13 x 11 = 143, remainder = 7
154 mod 13: 13 x 11 = 143, remainder = 11
158 mod 13: 13 x 12 = 156, remainder = 2

Product mod 13 = (3 x 7 x 11 x 2) mod 13

3 x 7 = 21, 21 mod 13 = 8
8 x 11 = 88, 88 mod 13 = 10  (13 x 6 = 78, 88-78 = 10)
10 x 2 = 20, 20 mod 13 = 7

Answer: 7