Playing with Numbers Test - 2

The option (C) is correct answer.

Number that is formed by the last two digits is 12, which is divisible by 4

Thus, 47,24,516 is divisible by 4.

We know that multiples of 12 are

12, 24, 36, 48, 60, 72, 84, 96, 108, 120.

Multiples of 16 are

16, 32, 48, 64, 80, 96, 112, 128, 144, 160 ….

Hence, the first two common multiples of 12 and 16 are 48 and 96.

We know that

The first bus stop at which both of them will stop = LCM of 8 and 12

So the required LCM = 2 × 2 × 2 × 3 = 24

This, 24th building is the first bus stop at which both of them will stop.

9999 is the largest 4-digit number

We know that

Therefore, 9999 is the largest 4-digit number and can be expressed as 3 × 3 × 11 × 101.

A 18 and 24

We know that

1 × 18 = 18

3 × 6 = 18

Factors of 18 are 1,2, 3, 6, 9 and 18 .

We know that

1 × 24 = 24

2 × 12 = 24

Factors of 24 are 1,2,3,4,6,8,12  and 24.

Therefore, the common factors are 1,2,3 and 6.

Perfect number is a positive integer that is equal to the sum of its proper divisors

The option (B) is correct answer.

We know that the factors of 6 are 1, 2, 3.

So the sum of the factors of 6 = 1 + 2 + 3 = 6

Consider brand A contain 28 and brand B contain 42 number of biscuits

We know that equal number of biscuits can be found by taking LCM of the number of chocolates

So we get

LCM of 28 and 42 is

So the required LCM = 2 × 2 × 3 × 7 = 84

Hence, 84 number of biscuits of each kind should be purchased.

No. of packets of brand A that should be purchased = 84 ÷ 28 = 3

No. of packets of brand B that should be purchased = 84 ÷ 42 = 2

Therefore, the least number of boxes of 3 of first kind and 2 of second kind should be purchased.

Here , we find the LCM of 180 and 240

So the required LCM = 2 × 2 × 2 × 2 × 3 × 3 x 5 = 720

Therefore, the next heap which lies at the foot of a pole is 720 m far along the road.

147*5

Sum of digits = 1 + 4 + 7 + 5 = 17

We know 18 is the multiple of 9 which is greater than 17.

So the smallest required number = 18 – 17 = 1

Hence, the smallest required number is 1.

(i) Two prime numbers are always co-prime to each other.

For example: The numbers 7 and 11 are co-prime to each other.

(ii) One prime and one composite number are not always co-prime.

For example: The numbers 3 and 21 are not co-prime to each other.

(iii) Two composite numbers are not always co-prime to each other.

For example: 4 and 6 are not co-prime to each other.

The dimensions of corridor are

Length = 20 m 16 cm = 2016 cm

Breadth = 15 m 60 cm = 1560 cm

Least possible side of square tiles used = HCF of 2016 and 1560

We know that the prime factorization of

2016 = 2 × 2 × 2 × 2 × 2 × 3 × 3 × 7

1560 = 2 × 2 × 2 × 3 × 5 × 13

So HCF of 2016 and 1560 = 2 × 2 × 2 × 3 = 24

Hence, the least possible size of square tiles used is 24 cm.

We know that

No. of square tiles which is used to pave the rectangular corridor = Area of courtyard/ Area of tile

By substituting the values

No. of square tiles which is used to pave the rectangular corridor = (2016 × 1560)/ (24)2 = 5460

Therefore, the least possible number of such tiles is 5460.

Given 

425 people donated Rs. 1000 each = 425 x 1000. = 425000

338 people donated Rs. 500 each = 338 x 500 = 169000.

Number of people who donated Rs. 100 each = 1763 – (425 + 338) 

⇒ 1763 – 763 = 1000

1000 people donated Rs. 100 each = 1000 x 100 = 100000

Total money collected for charity = 425000 + 169000 + 100000

= 694000

We find out the LCM of 28, 42, 56

So the required LCM = 2 × 2 × 2 × 3 × 7 = 168

The smallest number which is exactly divisible by 28, 42 and 56 is 168

In order to get remainder as 3

Required smallest number = 168 + 3 = 171

Therefore, the smallest number which when divided by 28, 42 and 56 gives a remainder of 3 each time is 171.

It is given that

LCM of two numbers = 260

HCF of two numbers = 8

One of the number = 40

We know that

Product of two numbers = Product of HCF and LCM

So we get

40 × other number = 8 × 260

On further calculation

Other number = (8 × 260)/ 40 = 52

Hence, the other number is 52

The option (C) is correct answer.

We know that 263 = 1 × 263

263 has two factors, 1 and 263

Therefore, it is a prime number

We know that a natural number is divisible by 2 if 0, 2, 4, 6 or 8 are unit digits.

 784645

The units digit in 784645 is 5

Therefore, 784645 is not divisible by 2.

The option (B) is correct answer.

We know that LCM of two co-prime numbers is equal to their product.

Hence, LCM of ‘x’ and ‘y’ will be xy.

We know that

Hence, the prime factorization of 13915 is 5 × 11 × 11 × 23.

The option (C) is correct answer.

We know that the sum of the given digits = 7 + 4 = 11

Multiple of 3 greater than 11 is 12.

So we get 12 − 11 = 1

Hence, the required digit is 1.

The option (B) is correct answer.

We know that 9 = 3 ×3 × 1 and 10 = 2 × 5 × 1

Both 9 and 10 are composite numbers with common factor 1

Hence, 9 and 10 are co-primes