Playing With Numbers — Competency Focused Questions

A number N satisfies the following properties:

• N is a 4-digit number
• The sum of its digits equals 18.
• N is divisible by both 9 and 11.

Which one of the following is the smallest possible value of N?

1. 1980

2. 2079

3. 3036

4. 1089

Answer

A number is divisible by 9 only if the sum of its digits is divisible by 9. The digit sum is 18, which is divisible by 9, so 1980, 2079 and 1089 satisfy this (the digit sum of 3036 is 12, so 3036 is rejected).

A number is divisible by 11 only if the difference between the sum of its digits at odd places and even places is 0 or a multiple of 11.

For the smallest option, 1089: (1 + 8) - (0 + 9) = 9 - 9 = 0, which is divisible by 11. Also 1089 = 9 × 121 = 11 × 99, so it is divisible by both 9 and 11.

Since, 1089 is the smallest value satisfying all the conditions, thus the smallest possible value of N = 1089.

Hence, Option 4 is the correct option.

If we multiply 2 prime numbers, the product will always be:

1. a composite number

2. an even number

3. an odd number

4. a prime number

Answer

The product of any two primes is greater than each prime and is divisible by at least one of those primes besides 1 and itself; therefore it is composite.

Hence, Option 1 is the correct option.

The two numbers nearest to 10000 which are exactly divisible by each of 2, 3, 4, 5, 6 and 7 are:

1. 9660, 10080

2. 9320, 10080

3. 9660, 10060

4. 10340, 10080

Answer

A number divisible by each of 2, 3, 4, 5, 6 and 7 must be a multiple of their L.C.M.

The L.C.M. of 2, 3, 4, 5, 6 and 7 is 420.

On dividing 10000 by 420, we get 23 with a remainder, so the nearest multiples of 420 are 420 × 23 = 9660 (just below 10000) and 420 × 24 = 10080 (just above 10000).

Since these are the multiples of 420 closest to 10000, thus the two required numbers are 9660 and 10080.

Hence, Option 1 is the correct option.

The H.C.F. and L.C.M. of two numbers are 13 and 1989 respectively. If one of the numbers is 117, the sum of both the numbers is:

1. 119

2. 221

3. 338

4. 439

Answer

We know that,

The other number = $\dfrac{\text{H.C.F.} \times \text{L.C.M.}}{\text{given number}}$ = $\dfrac{13 \times 1989}{117} = \dfrac{25857}{117}$ = 221

Since the other number is 221, thus the sum of both the numbers = 117 + 221 = 338.

Hence, Option 3 is the correct option.

Which of the following statements is true?

1. 1 is the smallest prime number.

2. If two numbers are co-primes, then at least one of them must be a prime number.

3. If a number is prime, it must be odd.

4. Two consecutive odd prime numbers are always twin primes.

Answer

Checking each statement:

1. 1 is not a prime number; the smallest prime number is 2. So this statement is false.

2. 8 and 9 are co-prime, yet both are composite. So at least one of them need not be prime, and this statement is false.

3. 2 is a prime number that is even. So a prime number need not be odd, and this statement is false.

4. Two consecutive odd numbers differ by 2; if both are prime, they form a twin-prime pair. So this statement is true.

Since only the fourth statement holds, thus the true statement is that two consecutive odd prime numbers are always twin primes.

Hence, Option 4 is the correct option.

You have to make a 3-digit number using the digits 0, 4, 5. The number you form will never be:

1. prime

2. composite

3. odd

4. even

Answer

The sum of the digits 0, 4 and 5 is 0 + 4 + 5 = 9, which is divisible by 3. So every 3-digit number formed using these digits is divisible by 3, and hence is composite.

(For example, 405 is odd, 450 is even and 504 is composite, so the number can be odd, even or composite.)

Since every such number is always divisible by 3, thus the number formed will never be a prime number.

Hence, Option 1 is the correct option.

P = L.C.M. of 2nd multiple of 9 and 6th multiple of 6.

Q = H.C.F. of 60 and 84.

L.C.M. of P and Q is:

1. 18

2. 24

3. 42

4. 36

Answer

The 2nd multiple of 9 is 9 × 2 = 18 and the 6th multiple of 6 is 6 × 6 = 36.

So, P = L.C.M. of 18 and 36 = 36.

Also, 60 = 2 × 2 × 3 × 5 and 84 = 2 × 2 × 3 × 7, so Q = H.C.F. of 60 and 84 = 2 × 2 × 3 = 12.

Since P = 36 and Q = 12, thus the L.C.M. of P and Q = L.C.M. of 36 and 12 = 36.

Hence, Option 4 is the correct option.

The product of three consecutive numbers is always divisible by:

1. 4

2. 6

3. 12

4. 24

Answer

Among any three consecutive numbers, at least one is divisible by 2 and exactly one is divisible by 3. So their product is always divisible by 2 × 3 = 6.

It need not be divisible by 4, 12 or 24; for example, 1 × 2 × 3 = 6 is divisible by 6 but not by 4, 12 or 24.

Since the product always contains the factors 2 and 3, thus the product of three consecutive numbers is always divisible by 6.

Hence, Option 2 is the correct option.

The students in a class can be divided into groups of 2, 3, 5 and 6. What is the least number of students this class can have?

1. 40

2. 30

3. 35

4. 42

Answer

The least number that can be divided exactly into groups of 2, 3, 5 and 6 is the L.C.M. of these numbers.

The L.C.M. of 2, 3, 5 and 6 = 30.

Since the least such number is the L.C.M. of the group sizes, thus the least number of students the class can have = 30.

Hence, Option 2 is the correct option.

Find the value of a + b + c, if 373a is divisible by 9, 473b is divisible by 11 and 371c is divisible by 6.

1. 7

2. 6

3. 0

4. 9

Answer

For 373a to be divisible by 9, the sum of its digits 3 + 7 + 3 + a = 13 + a must be divisible by 9. This gives a = 5.

For 473b to be divisible by 11, the difference between the sum of digits at odd places and even places, (b + 7) - (3 + 4) = b, must be 0 or a multiple of 11. This gives b = 0.

For 371c to be divisible by 6, it must be divisible by both 2 and 3. For divisibility by 2, c must be even; for divisibility by 3, the digit sum 3 + 7 + 1 + c = 11 + c must be divisible by 3. The even digit satisfying this is c = 4.

Since a = 5, b = 0 and c = 4, thus a + b + c = 5 + 0 + 4 = 9.

Hence, Option 4 is the correct option.

Read the following statements and choose the correct option:

Statement-1: A number for which the sum of all its factors is equal to twice the number is called a perfect number.

Statement-2: The product of two prime numbers is always odd.

1. Both the statements are true.

2. Statement-1 is true but Statement-2 is false.

3. Statement-1 is false but Statement-2 is true.

4. Both the statements are false.

Answer

Statement-1 is the correct definition of a perfect number. For example, the factors of 6 are 1, 2, 3 and 6, and their sum 1 + 2 + 3 + 6 = 12 = 2 × 6. So, statement-1 is true.

The product of two prime numbers is not always odd; for example, 2 × 3 = 6 is even. So, statement-2 is false.

∴ Statement-1 is true but Statement-2 is false.

Hence, Option 2 is the correct option.