Playing with Numbers Test - 5

A four digit number abcd can be expressed as:

abcd = 1000a + 100b + 10c + d

= (1001a + 99b + 11c) − (a − b + c − d)

= 11 (91a + 9b + c) + [(b + d) − (a + c)]

∴X = 11 (91a + 9b + c)

Thus, X is always divisible by 11.

The correct answer is D.

We know that if a number is divisible by two co-prime numbers it is also divisible by their product.

Now,

45 = 5 × 9

Here, 5 and 9 are co-prime numbers.

879xy is divisible by 45. Therefore, it must be divisible by both 5 and 9.

A number is divisible by 5 if its units digit is 0 or 5, and it is divisible by 9 if the sum of its digits is a multiple of 9.

∴ y = 0 or 5

If y = 0, then the number becomes 879x0.

Now, 879x0 is divisible by 9 if 8 + 7 + 9 + x + 0 = 24 + x is divisible by 9.

∴ 24 + x = 27, 36 …

Also, x is a single-digit number.

∴ x = 3

⇒ x² − 2y² = 9 − 2 × 0² = 9

If y = 5, the number becomes 879x5.

879x5 is divisible by 9 if 8 + 7 + 9 + x + 5 = 29 + x is divisible by 9.

∴ 29 + x = 36, 45 …

⇒ x = 7

∴ x² − 2y² = − 1

Thus, the possible values of the expression x² − 2y² are 9 and −1.

The correct answer is A.

We know that if a number is divisible by two prime numbers, then it is also divisible by their product.

Now, 55 = 5 × 11, where 5 and 11 are prime numbers.

x313y is divisible by 55. Therefore, it must be divisible by 5 and 11.

A number is divisible by 5, if the digit at its unit's place is either 0 or 5, and a number is divisible by 11 if the difference between the sum of its digits at odd places and the sum of its digits at even places is either 0 or a multiple of 11.

Therefore, y = 0 or 5.

If y = 5, then the number becomes x3135.

Sum of digits at odd places = 6 + x

Sum of digits at even places = 6

∴ Difference = 6 + x − 6 = x

Since the given number is a five digit number and x is a single digit number, x cannot be 0 and cannot be a multiple of 11.

∴y ≠ 5

If y = 0, then the number becomes x3130.

Sum of digits at odd places = x + 1

Sum of digits at even places = 6

∴ Difference = 6 − (x + 1) = 5 − x

Since x is a single digit number, the only possible value of x is 5.

∴ x = 5 and y = 0

⇒ x² + y² + x = 5² + 0² + 5 = 30

The correct answer is C.

The given four-digit number is 103x.

A number is divisible by 3 if the sum of its digits is divisible by 3.

Now, 103x = 1 + 0 + 3 + x = 4 + x

Therefore, 103x is divisible by 3 if 4 + x is divisible by 3.

Since x is a single digit number,

4 + x = 6 or 9 or 12

⇒ x = 2 or 5 or 8

Therefore, the possible values of x are 2, 5, 8.

Thus, x can have three possible values.

Hence, the correct answer is option C.

A number is divisible by 5 if its units digit is either 0 or 5 and a it is divisible by 2 if its units digit is 0 or any even number.

The number leaves the remainder 4 when divided by 5.

∴y = 0 + 4 = 4 or 5 + 4 = 9

Now, the number leaves no remainder when divided by 2.

∴y = 4

Therefore, the number is 1x794.

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

Sum of digits at even places = 9 + x

Sum of digits at odd places = 12

∴ Difference = 12 − (9 + x) or 9 + x − 12

= 3 − x or x − 3

Since x is a single-digit number, 3 − x or x − 3 cannot be a multiple of 11.

∴ 3 − x = 0 or x − 3 = 0

⇒ x = 3

∴ xy = 3 × 4 = 12

Thus, the product of x and y is 12.

The correct answer is C.

We know that if a number is divisible by two prime numbers, it is also divisible by their product.

Now, 33 = 3 × 11

Here, 3 and 11 are prime numbers.

2xy02 is divisible by 33. Therefore, it must be divisible by 3 and 11.

A number is divisible by 3 if the sum of its digits is divisible by 3. A number is divisible by 11 if the difference between the sum of the digits at odd places and the sum of the digits at even places of the number is either 0 or a multiple of 11.

Sum of digits = 2 + x + y + 0 + 2 = 4 + x + y

Now, 2xy02 is divisible by 3 if 4 + x + y is divisible by 3.

∴ 4 + x + y = 9, 12, 15 …

⇒ x + y = 5, 8, 11 … … (1)

Sum of the digits at odd places = 4 + y

Sum of the digits at even places = 0 + x

∴ Difference = 4 + y − x [y > x]

Now, 2xy02 is divisible by 11 if 4 + y − x is either 0 or a multiple of 11.

∴ 4 + y − x = 0, 11, 22 …

y − x > 0 ⇒ 4 + y − x ≠ 0

∴ 4 + y − x = 11, 22, 33 …

⇒ y − x = 7, 18, 29 … … (2)

Since y and x are single digit numbers, therefore, y − x = 7.

It can be seen that the only possible values of x and y that satisfy equations (1) and (2) are y = 9 and x = 2.

∴ xy − x − y = 9 × 2 − 2 − 9

= 18 − 2 − 9

= 7

The correct answer is B.

We know that a number is divisible by 3, if the sum of its digits is divisible by 3.

It is given that the numbers 311x2 and 42y5 are divisible by 3.

Sum of its digits = 3 + 1 + 1 + x + 2= 7 + x

Thus, 7 + x is divisible by 3.

⇒ 7 + x = 3, 6, 9, 12 …

Since x is the smallest digit, x = 2.

Sum of the digits of the number 42y5 = 4 + 2 + y + 5 = 11 + y.

11 + y is divisible by 3.

⇒ 11 + y = 3, 6, 9, 12 …

Since y is the smallest digit, y = 1.

Thus, the required difference is 2 −1 = 1.

The correct answer is A.

We know that 18 = 2 × 9, where 2 and 9 are co-prime.

It is known that if a number is divisible by two co-prime numbers, then it is also divisible by their product and vice-versa.

Therefore, the given number is divisible by both 2 and 9.

It is clearly seen that for any value of digit p, the number 123p2 is divisible by 2 as the unit's digit is even.

As the given number is divisible by 9, the sum of its digits is divisible by 9.

Sum of digits = 1 + 2 + 3 + p + 2 = 8 + p

∴ 8 + p is divisible by 9.

⇒ 8 + p = 9, 18, …

Since p is a single digit number, the only possible value of p is 1.

Thus, the value of digit p is 1.

The correct answer is A.

It can be seen that the four digit number abcd has a thousands, b hundreds, c tens, and d ones.

Thus, the number abcd can be written in the general form as

abcd =1000 × a + 100 × b + 10 × c + d.

The correct answer is A.

A three digit numberabc can be written as:

abc = 100a + 10b + c

Similarly, bca and cab can be written as:

bca = 100b + 10c + a

cab = 100c + 10a + b

∴abc + bca + cab = (100a + 10a + a) + (100b + 10b + b) + (100c + 10c + c)

= 111a + 111b + 111c

= 111 (a + b + c)

= 3 × 37 (a + b + c)

The correct answer is B.

Number A can be obtained by simplifying the general form:

A = 10000 × w + 1000 × x + 100 × y + z

= 10000 × w + 1000 × x + 100 × y +10 × 0 + z

= w0000 + x000 + y00 + z

= wxy0z

Thus, the value of number A is wxy0z.

Hence, the correct answer is option C.

A three digit number abc can be written as:

abc = 100a + 10b + c

The number formed by reversing its digits is cba, which can be written as:

cba = 100c + 10b + a

If a > c, then the difference between the numbers is

(100a + 10b + c) − (100c + 10b + a)

= 100a + 10b + c − 100c − 10b − a

= 99a − 99c

= 99 (a − c)

If c > a, then the difference between the numbers is

(100c + 10b + a) − (100a + 10b + c)

= 99c − 99a

= 99(c − a)

Thus, the difference can be written as 99(a − c) or 99 (c − a).

The correct answer is B.

A number is divisible by 5 if the digit at its units place is either 0 or 5.

A number is divisible by 2 if the digit at its units place is 0 or any even number.

Now, the number leaves 2 as remainder when divided by 5.

Therefore, the digit at its units place is either 0 + 2 = 2 or 5 + 2 = 7.

It is given that the number leaves no remainder when divided by 2.

Therefore, the units digit of the number is even i.e. 2.

Thus, the units digit of the number is 2.

The correct answer is B.