Number System Test - 1

Let us assume that the number with which Anita has to perform the multiplication is 'x'.

Instead of finding 35x, she calculated 53x.

The difference = 53x - 35x = 18x = 540

Therefore, x = 540/18 = 30

So, the new product = 30 x 53 = 1590.

Any factor of this number should be of the form 2^a × 3^b × 5^c
For the factor to be a perfect square a, b, c has to be even.
⇒ a can take values 0, 2, 4 
⇒ b can take values 0, 2, 4, 6
⇒ c can take values 0, 2
So, total number of perfect squares = 3 × 4 × 2 = 24

For number of zeroes we must count number of 2 and 5 in prime numbers below 100.
We have just 1 such pair of 2 and 5.
Hence we have only 1 zero.

We have a + c = e so possible summation 6+4=10 or 4+2 = 6.
Also b = 2d so possible values  4 = 2 * 2 or 10 = 5 * 2.
So considering both we have b = 10 , d = 5, a= 4 ,c = 2, e = 6.
Hence the correct option is B .

The sum of the first 100 natural numbers is:

=  (n * (n + 1)) / 2
=  (100 * 101) / 2
=  50 * 101

101 is an odd number and 50 is divisible by 2.
Hence, 50 * 101 will be divisible by 2.

Let the 4 digit no. be xyzw.
According to given conditions we have x + y = z + w, x + w = z, y + w = 2x + 2z.
With help of these equations, we deduce that y = 2w, z = 5x.
Now the minimum value x can take is 1 so z = 5 and the no. is 1854, which satisfies all the conditions. Hence option A.

If the number of pages is 44, the sum will be 44*45/2 = 22*45 = 990
So, the number 10 was added twice.

The difference of the numbers = 34041 – 32506 = 1535
The number that divides both these numbers must be a factor of 1535.
307 is the only 3 digit integer that divides 1535.

In this type of question, We need to find out the LCM of the given numbers.
LCM of 12, 15, 18 and 20:

⇒ 12 = 2 x 2 x 3
⇒ 15 = 3 x 5
⇒ 18 = 2 x 3 x 3
⇒ 20 = 2 x 2 x 5

∴ LCM = 2 x 2 x 3 x 5 x 3
Since the soldiers are in the form of a solid square. Hence, LCM must be a perfect square.
To make the LCM a perfect square, We have to multiply it by 5, hence, the required number of soldiers: 

= 2 x 2 x 3 x 3 x 5 x 5
= 900

The factors of 1080 which are perfect square:

1080 → 2³ × 3³ × 5

For, a number to be a perfect square, all the powers of numbers should be even number.

Power of 2 → 0 or 2
Power of 3 → 0 or 2
Power of 5 → 0 

So, the factors which are perfect square are 1, 4, 9, 36.
Hence, Option B is correct.

• Different possibilities for the number of pencils = 12 or 13.
• Since it cannot be divided into his 4 brothers and sisters, it has to be 13.
• The number of erasers should be less than the number of pencils and greater than or equal to 11. So the number of erasers can be 11 or 12.
• If the number of erasers is 12, then the number of pens = 38 - 13 - 12 = 13, which is not possible as the number of pens should be more than the number of pencils.
• So the number of erasers = 11 and therefore the number of pens = 14 

The size of each row would be the HCF of 363, 429 and 693. Difference between 363 and 429 =66.

Factors of 66 are 66, 33, 22, 11, 6, 3, 2, 1.

66 need not to be checked as it is even and 363 is odd. 33 divides 363, hence would automatically divide

429 and also divides 693. Hence, 33 is the correct answer for the size of each row.

For how many rows would be required we need to follow the following process:

Minimum number of rows required = 363/33 + 429/33 + 693/33 = 11 + 13 + 21 = 45 rows.

Therefore, the correct answer is A

• There are several rational numbers between 4 and 5. The numbers are between 16/4 and 20/4. 
• Therefore, the answer is c, that is, 17 / 4, 18 / 4, 19 / 4.

Solve this question step by step:

1. Any number formed by this method is clearly divisible by 3.
2. Since it needs to be a square, it should be divisible by (3)^[2*k]. k varies over the natural numbers.
3. Now consider the original number. It has hundred 1’s, hundred 2’s and hundred 3’s. Sum of these digits is 600. This is not divisible by 9. Hence number is not divisible by 9.
4. If a number is divisible by (3)^[2*k], it is divisible by 3^k.
5. This number is not divisible by 3^k for any k > 1. 

Hence it is not a perfect square for any arrangement.

Remainder = (73 x 75 x 78 x 57 x 197 x 37) / 34
= (5 x 7 x 10 x 23 x 27 x 3) / 34

(We have taken individual remainder, which means if 73 is divided by 34 individually, it will give remainder 5, 75 divided 34 gives remainder 7 and so on.)

= (5 x 7 x 10 x 23 x 27 x 3) / 34
= (35 x 30 x 23 x 27) / 34
= (1 x (-4) x (-11) x (-7)) / 34

(We have taken here negative as well as positive remainder at the same time. When 30 divided by 34 it will give either positive remainder 30 or negative remainder -4. We can use any one of the negative or positive remainders at any time.)

= (1 x -4 x -11 x -7) / 34
= (28 x -11) / 34 
= (-6 x -11) / 34
= 66 / 34

∴  R = 32

Find the difference between both 34041 and 32506.

34041 - 32506 = 1535

So, N should be a factor of 1535. Factors of 1535:

1, 5, 307, 1535

Here, 307 is a three-digit factor of 1535. So, N = 307.
34041 and 32506 divided by 307 leaves remainder 271. So, 307 is the answer.

Since after division of a number successively by 3, 4 and 7, the remainders obtained are 2, 1 and 4 respectively, the number is of form ((((4*4)+1)*3)+2)k = 53K.

Let k = 1; the number becomes 53

If it is divided by 84, the remainder is 53.

Therefore, the correct answer is Option D.

We are provided with the equation (32) + (24) = (100). Let us assume our base be 'b'
Then,we can say:

⇒ 32 = 3 x b¹ + 2 x b⁰ = 3b+2
⇒ 24 = 2 x b¹+ 4 x b⁰ = 2b+4
⇒ 100 = 1 x b² + 0 x b¹ + 0 x b⁰ = b²

Now, according to our question:

⇒ 32 + 24=100
⇒ (3b + 2) + (2b + 4) = (b²)
⇒ 5b + 6 = b²
⇒ b² - 5b - 6 = 0
⇒ b² - 6b + b - 6 = 0
⇒ b(b - 6) + 1(b - 6) = 0
⇒ (b - 6) * (b + 1) = 0
⇒ b = 6,- 1

Base can't be negative. Hence b = 6.
∴ Base assumed in the asked question must be 6.

24 = 8 × 3
Therefore, we need to find the highest power of 8 and 3 in 150!
8 = 2³
Highest power of 8 in 150! is:

= [(150 / 2) + (150 / 4) + (150 / 8) + (150 / 16) + (150 / 32) + (150 / 64) +(150 / 128)] / 3
= 48

Highest power of 3 in 150! is:

= [150 / 3] + [150 / 9] + [150 / 27] + [150 / 81]
= 72

As the powers of 8 are less, powers of 24 in 150! = 48

Factors: Beauty of the number 1001. This number is not prime, is a product of three distinct primes and does wonderful things to three-digit numbers when multiplied to them.

To start with ‘abcabc’ = ‘abc’ * 1001 or abc * 7 * 11 * 13 (This is a critical idea to remember).

‘abc’ has only two factors. Or, ‘abc’ has to be prime. Only a prime number can have exactly two factors. (This is in fact the definition of a prime number)

So, ‘abcabc’ is a number like 101101 or 103103.

’abcabc’ can be broken as ‘abc’ * 7 * 11 * 13. Or, a p * 7 * 11 * 13 where p is a prime.

As we have already seen, any number of the form paqbrc will have (a + 1) (b + 1) (c + 1) factors, where p, q, r are prime.

So, p * 7 * 11 * 13 will have = (1 + 1) * (1 + 1) * (1 + 1) * (1 + 1) = 16 factors

The question is "How many factors does the 6-digit number ‘abcabc’ have?"

Hence the answer is 16 factors.
Choice A is the correct answer.