Test: Number Play - 2 - Class 6 MCQ

Solution:

A supercell is a number greater than all its neighbouring cells. In the grid [45, 78, 92, 31, 60], we identify the supercell as follows:

• For 45: Neighbour is 78 (right). Not a supercell.
• For 78: Neighbours are 45 (left) and 92 (right). Not a supercell.
• For 92: Neighbours are 78 (left) and 31 (right). This is a supercell.
• For 31: Neighbours are 92 (left) and 60 (right). Not a supercell.
• For 20: Neighbour is 31 (left). Not a supercell.

Therefore, the only supercell in the grid is 92.

A 3-digit palindromic number has the form ABA, where:

• The first digit (A) and the third digit (A) are the same.
• The middle digit (B) can be any digit.

Step-by-Step Analysis:

We are given the digits 1, 2, and 3, and we need to form a 3-digit palindromic number.

1. The first and third digits (A): There are 3 possible choices for A: 1, 2, or 3.
2. The middle digit (B): The middle digit can be any of the three digits: 1, 2, or 3. Thus, there are 3 possible choices for B.

The total number of possible 3-digit palindromic numbers is:

3 (choices for A) × 3 (choices for B) = 9

There are 9 possible 3-digit palindromic numbers that can be formed using the digits 1, 2, and 3.

Let's find how many times the digit '7' appears in the numbers from 1 to 100.

Step-by-Step Solution:

Step 1: Count the '7's in the Tens Place

The tens place is the second digit of the number. We need to find the numbers that have '7' in the tens place. These numbers are:

• 70, 71, 72, 73, 74, 75, 76, 77, 78, 79

So, there are 10 numbers that have '7' in the tens place.

Step 2: Count the '7's in the Ones Place

The ones place is the last digit of the number. We need to find the numbers that end with '7'. These numbers are:

• 7, 17, 27, 37, 47, 57, 67, 77, 87, 97

So, there are 10 numbers that have '7' in the ones place.

Step 3: Total Count of '7's

From the tens place, we have 10 '7's (from 70 to 79).
From the ones place, we have 10 '7's (from 7, 17, 27, etc.).
So, the total number of '7's is:

10 + 10 = 20

The digit '7' appears 20 times when writing all the numbers from 1 to 100.

The correct answer is C: 20.

The problem asks for the smallest 4-digit number that, when reversed, gives a 4-digit number.

1. Reversing a 4-digit number means swapping its digits in reverse order.

2. We are looking for a number where its reverse is also a 4-digit number.

Let's examine the given options:

• A: 1000: Reversing 1000 gives 0001, which is not a 4-digit number.

• B: 1001: Reversing 1001 gives 1001, which is a 4-digit number.

• C: 1100: Reversing 1100 gives 0011, which is not a 4-digit number.

• D: 1010: Reversing 1010 gives 0101, which is not a 4-digit number.

Thus, B: 1001 is our required answer.

The Kaprekar constant for 4-digit numbers is 6174.

Here's the process to find the Kaprekar constant for any 4-digit number:

1. Start with any 4-digit number, using at least two distinct digits.
2. Rearrange the digits to form the largest and smallest possible numbers.
3. Subtract the smaller number from the larger number.
4. Repeat the process with the result.
5. Eventually, you will always reach 6174, which is the Kaprekar constant.

Let's start with the number 3524:

• Rearrange the digits to form the largest and smallest numbers: 5432 and 2345.
• Subtract the smaller number from the larger number: 5432 - 2345 = 3087.
• Repeat the process with 3087: 8730 - 0378 = 8352.
• Continue until you reach 6174.

The Kaprekar constant for 4-digit numbers is 6174.

A palindrome is a number (or word) that reads the same forwards and backwards.

Let’s check each option to see if it’s a palindrome:

• A: 34544: When we reverse it, we get 44543, which is not the same as 34544. So, it’s not a palindrome.

• B: 12321: When we reverse it, we get 12321, which is the same as the original number. So, this is a palindrome.

• C: 65457: When we reverse it, we get 75456, which is not the same as 65457. So, it’s not a palindrome.

• D: 78998: When we reverse it, we get 88997, which is not the same as 78998. So, it’s not a palindrome.

The number that is a palindrome is B: 12321.

So, the correct answer is B: 12321.

A 5-digit palindrome has the form abcba. Here a=1 and c=2·b. Since c must be a digit (0–9), 2·b≤9 gives b=0 and c=0. Thus the number is 1 0 0 0 1, i.e. 10001.

• Arrange numbers by size from left to right on the line.
• 2180 is smaller than 3050, so it goes first.
• 3050 is less than 9950, so it follows 2180.
• 9950 is the largest, so it sits at the right end.
• The correct ascending sequence is 2180, 3050, 9950, which is option A.

The sum of the digits of 176 can be calculated as follows:

• Identify the digits: 1, 7, and 6.
• Add them together: 1 + 7 + 6.
• The result is 14.

To form the largest number using the digits 6382, we need to arrange the digits in descending order (from highest to lowest).

The digits are: 6, 3, 8, 2

Arranging them in descending order gives: 8, 6, 3, 2

Thus, the largest number that can be formed is 8632.

Therefore, the correct answer is A: 8632.