Number System Test

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Number System Test - 6

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Weekly Quiz Competition

Question 1
3 / -1

0.xyxyxyxy.... + 0.yxyxyxyx = 4/3, if x and y are natural numbers what would be the value of x + y?

A
9

B
11

C
12

D
15

Solution

= x + y = 12
Question 2
3 / -1

The sum of first five, 3 digit prime numbers is:

A
547

B
533

C
529

D
555

Solution

► First 5, 3 digit prime numbers are listed below:

► Their Sum is 533
Question 3
3 / -1

If a, (a+2), (a+4) are all prime numbers, how many such values can 'a' take?

A
None

B
1

C
Infinite

D
None of the Above

Solution

► The given terms are: a, a + 2, a + 4
► For Less than 5, We know of such a set: 3, 5 and 7
► For Prime numbers Greater than 5 we have two cases
► Case 1 : The Prime number on division by 6 gives a remainder of 1
• Let us assume a is a prime number
• So a = 6p + 1
• a + 2 = 6p +3. (This will be divisible by 3. Thus not a prime)
• a + 4 = 6p + 7
► Case 2 : The Prime number on division by 6 gives a remainder of 5
• Let us assume a is a prime number
• So a = 6p + 5
• a + 2 = 6p + 7
• a + 4 = 6p + 9 (This will be divisible by 3. Thus not a prime.)
► Hence there is only one such set.
Question 4
3 / -1

What is the sum of all the negative two digit prime numbers?

A
1043

B
-1043

C
-1060

D
None of the above

Solution

► Prime numbers is a property of natural numbers, There are no negative prime numbers.
Question 5
3 / -1

What is the negative of sum of all the two digit prime numbers?

A
1060

B
-1043

C
-1060

D
None of the above

Solution

► All the two digit Prime numbers are listed below:

►The sum is 1043
►The negative of this sum is - 1043
Question 6
3 / -1

For the equation (2a + 3b + 9c) to be even, we must have

A
b should be even

B
c should be even

C
b and c should be even

D
b or c should be even

Solution

► Option A : b is Even
⇒ 2a + 3b + 9c = 2 x any number + 3 x Even + 9 x any number = Even + Even + Odd/Even
⇒ This does not give a sure shot result.
► Option B : c is Even
⇒ 2a + 3b + 9c = 2 x any number + 3 x any number + 9 x Even = Even + Odd/Even + Even
⇒ This does not give a sure shot result.
► Option D : Either of the two (b or c) is Even.
⇒ 2a + 3b + 9c = 2 x any number + 3 x Even + 9 x any number = Even + Even + Odd/Even
⇒ 2a + 3b + 9c = 2 x any number + 3 x any number + 9 x Even = Even + Odd/Even + Even
⇒ This does not give a sure shot result.
► Option C = Both b and c are Even
⇒ 2a + 3b + 9c = 2 x any number + 3 x Even + 9 x Even = Even + Even + Even = Even
⇒ So only option C gives an even response.
Question 7
3 / -1

What is the difference between the highest and smallest 3 digit prime number?

A
896

B
998

C
990

D
888

Solution

► Highest 3 digit prime number = 997
► Smallest 3 digit prime number = 101
► Difference = 997 - 101 = 896
Question 8
3 / -1

Suppose the sum of n consecutive integers is x+(x+1) +(x+2) +(x+3)+...+(x+(n-1)) =1000, then which of the following cannot be true about the number of terms n?

A
The number of terms can be 16

B
The number of terms can be 5

C
The number of terms can be 25

D
The number of terms can be 20

Solution

Solve this question through the options. For n terms being 16 (option 1), we would need an AP with 16 terms and common difference 1, that would add up to 1000. Since, the average value of a term of this AP turns out to be 1000÷16=62.5, we can create a 16 term AP as 55,56,57....62,63,64...70 that adds up to 1000. Hence, a series of16 consecutive terms is possible. Likewise, a series of 5 terms gives the average as 200 & the 5 terms can be taken as 198,199,200,201,202. It is similarly possible for 25 terms with an average of 40 but is not possible for 20 terms with an average of 50.
Hence, option (d) is correct.
Question 9
3 / -1

What would the remainder when the square of the largest 4 digit prime number is divided by 6?

A
1

B
5

C
0

D
None of the above

Solution

► It is +1 with exceptions of prime numbers below 6 obviously.
► Any prime number above 6 is of the form:
6k + 1 or 6k - 1 ( A lot of people don't know this!))
► So therefore squares is of the form (6k ± 1)². Expand this and clearly the only term which is not a multiple of 6 will be 1. Which again makes it of the form 6K + 1( now only + because we squared the whole term, Remember?)!.
► So therefore the remainder is of the form +1.
► So for every prime no. above 6, the remainder 1.
So, for 2, it is 4
For 3, it is 3.
For 5, it is 1.
► For every other prime no. Square, It’s 1.
Question 10
3 / -1

Let us consider a fraction whose denominator is smaller than the square of the numerator by unity. If we add 2 to the numerator and the denominator, the fraction will exceed1/3; now if we subtract 3 from the numerator and the denominator, the fraction remains positive but smaller than 1/10. Find the fraction.

A
3/8

B
4/15

C
5/24

D
None

Solution

According to the question the given fraction is N2−1N

⟹ Adding 2 to Numerator and denominator gives N2+1N+2>31

⟹N2−3N−5<0

⟹Now subtracting 3 gives N2−2N−1<101

⟹N2−10N+8>0

There is no such Integer value of N that can satisfy the conditions

So, the correct answer is D
Question 11
3 / -1

If we divide a two-digit number by the sum of its digits, we get 4 as a quotient and 3 as a remainder. Now if we divide that two-digit number by the product of its digits, we get 3 as a quotient and 5 as a remainder and the two-digit number.

A
25

B
23

C
30

D
27

Solution

Let n(a,b) is two digit number.
n(a,b)=10a+b
Given sum of its digitsnumber will give 4 as quotient and 3 as remqinder.
n(a,b)=4(a+b)+3
10a+b=4a+4b+3
6a−3b=3⇒2a=1+b
n(a,b)=3(ab)+5
10a+b=3ab+5
10a+2a−1=3a(2a−1)+5
12a−1=6a2−3a+5
6a2−15a+6=0
2a2−5a+2=0
(2a−1)(a−2)=0
a=2(∵a=21)
∴b=2a−1=3
23 is required number.
So, the correct option is B.
Question 12
3 / -1

What two-digit number is less than the sum of the squares of its digits by 11 and exceed their doubled product by 5?

A
15 and 95

B
95

C
Both (a) and (b)

D
15,95,12345

Solution

Let n(a,b) be two digit number.
n(a,b)=10a+b
Given that,   n(a,b)+11=a2+b2
10a+b+11=a2+b2→(1)
n(a,b)=2ab+5
10a+b=2ab+5→(2)
(1)−(2)⇒11=a2+b2−2ab−5
(a−b)2=16
a−b=±4⇒a=b+4(or)a=b−4
a=b+4,
10b+40+b=2b(b+4)+5
2b2−3b−35=0
(2b+7)(b−5)=0
∴  b=5,a=9
a=b−4
10b−40+b=2b2−8b+5
2b2−19b+45=0
(2b−9)(b−5)=0
∴  b=5,a=1

So, the correct option is A
Question 13
3 / -1

The last three - digits of the multiplication 12345 × 54321 will be

A
745

B
845

C
945

D
645

Solution

Divide the given expression by 1000 and find the remainder to get the answer. (12345 × 54321) ÷ 1000 = (2469 × 54321) ÷ 200 → gives a remainder of (69 × 121) ÷ 200 = 8349 ÷ 200 → gives us a remainder of 149. Thus, the remainder would be 149 × 5 = 745. Hence, the last three digits would be 745.
So, the correct answer is A
Question 14
3 / -1

Find the gcd (111....11 hundred ones; 11....11 sixty ones).

A
111....forty ones

B
111.....twenty five ones

C

D
111.... sixty ones

Solution

The correct option is C 111..... twenty ones
The required GCD would be 1111...111 twenty ones.
Question 15
3 / -1

The remainder when 1010+10100+101000+....+1010000000000 is divided by 7 is

A
1

B
2

C
5

D
7

Solution

The remainder of (1010+10100+101000+....+1010000000000)÷7→(310+3100+31000+.....+310000000000)÷7→(34+34+34+34+34+34+34+34+34+34)÷7=Remainder of 40÷7→5.
So, the correct option is C

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Re-Attempt Weekly Quiz Competition

Number System Test - 6

Result

Number System Test - 6

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SHARING IS CARING

Question 1

9

11

12

15

Solution

Question 2

547

533

529

555

Solution

Question 3

None

1

Infinite

None of the Above

Solution

Question 4

1043

-1043

-1060

None of the above

Solution

Question 5

1060

-1043

-1060

None of the above

Solution

Question 6

b should be even

c should be even

b and c should be even

b or c should be even

Solution

Question 7

896

998

990

888

Solution

Question 8

The number of terms can be 16

The number of terms can be 5

The number of terms can be 25

The number of terms can be 20

Solution

Question 9

1

5

0

None of the above

Solution

Question 10

3/8

4/15

5/24

None

Solution

Question 11

25

23

30

27

Solution

Question 12

15 and 95

95

Both (a) and (b)

15,95,12345

Solution