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Question

Consider two-digit numbers which remain the same when the digits interchange their positions. What is the average of such two-digit numbers?

This question was previously asked in
CDS I 2019 Elementary Mathematics Previous Year Paper (03-Feb-2019)
The correct answer is

55

Finding Two-Digit Numbers with Interchangeable Digits

The question asks us to find the average of two-digit numbers that remain the same when their digits are interchanged. Let's first identify these special two-digit numbers.

A two-digit number can be represented as \(10 \times a + b\), where \(a\) is the tens digit and \(b\) is the units digit. For a two-digit number, \(a\) must be between 1 and 9 (inclusive), and \(b\) must be between 0 and 9 (inclusive).

When the digits are interchanged, the new number becomes \(10 \times b + a\).

The condition given is that the original number and the number with interchanged digits are the same:

\(\qquad 10a + b = 10b + a\)

Let's solve this equation to find the relationship between \(a\) and \(b\):

Subtract \(a\) from both sides:

\(\qquad 10a - a + b = 10b + a - a\)

\(\qquad 9a + b = 10b\)

Subtract \(b\) from both sides:

\(\qquad 9a + b - b = 10b - b\)

\(\qquad 9a = 9b\)

Divide both sides by 9:

\(\qquad \frac{9a}{9} = \frac{9b}{9}\)

\(\qquad a = b\)

This equation \(a = b\) tells us that the tens digit must be equal to the units digit for the two-digit number to remain the same when its digits are interchanged.

Since \(a\) is the tens digit of a two-digit number, \(a\) cannot be 0. The possible values for \(a\) (and thus \(b\)) are 1, 2, 3, 4, 5, 6, 7, 8, and 9.

The two-digit numbers where the digits are the same are:

  • When \(a=1, b=1\): The number is 11.
  • When \(a=2, b=2\): The number is 22.
  • When \(a=3, b=3\): The number is 33.
  • When \(a=4, b=4\): The number is 44.
  • When \(a=5, b=5\): The number is 55.
  • When \(a=6, b=6\): The number is 66.
  • When \(a=7, b=7\): The number is 77.
  • When \(a=8, b=8\): The number is 88.
  • When \(a=9, b=9\): The number is 99.

There are 9 such two-digit numbers.

Calculating the Average of These Numbers

To find the average of these two-digit numbers, we need to sum them up and divide by the count of the numbers.

The list of numbers is: 11, 22, 33, 44, 55, 66, 77, 88, 99.

The count of these numbers is 9.

The sum of these numbers is:

Sum \(= 11 + 22 + 33 + 44 + 55 + 66 + 77 + 88 + 99\)

We can factor out 11 from each term:

Sum \(= 11 \times (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9)\)

The sum of the first 9 natural numbers is given by the formula \(\frac{n(n+1)}{2}\), where \(n=9\).

Sum of \((1 + 2 + \dots + 9) = \frac{9 \times (9+1)}{2} = \frac{9 \times 10}{2} = \frac{90}{2} = 45\).

So, the sum of the two-digit numbers is:

Sum \(= 11 \times 45 = 495\)

Now, we calculate the average:

Average \(= \frac{\text{Sum of numbers}}{\text{Count of numbers}}\)

Average \(= \frac{495}{9}\)

Performing the division:

\(\qquad 495 \div 9 = 55\)

The average of the two-digit numbers which remain the same when the digits interchange their positions is 55.

Revision Table: Two-Digit Numbers and Average

Concept Description Application in this problem
Two-digit number A number with a tens digit (1-9) and a units digit (0-9). Represented as \(10a+b\). Identifying numbers where \(a,b\) are digits and \(a \ne 0\).
Interchanging Digits Swapping the tens and units digits. Number becomes \(10b+a\). Setting \(10a+b = 10b+a\) to find the condition.
Condition for equality \(a=b\) for the number to remain same. Finding all two-digit numbers with identical digits.
Average Sum of values divided by the count of values. Calculating \(\frac{\text{Sum}(11, 22, ..., 99)}{9}\).

Additional Information: Properties of Numbers

This problem touches upon basic number properties and arithmetic means (averages).

  • Numbers like 11, 22, 33 are sometimes called repdigits or repeating digits numbers. These are numbers formed by repeating a single digit.
  • Any two-digit number with identical digits is a multiple of 11. For example, \(11 = 11 \times 1\), \(22 = 11 \times 2\), ..., \(99 = 11 \times 9\).
  • The sum of an arithmetic progression can be calculated efficiently. In this case, the list 11, 22, ..., 99 is an arithmetic progression with the first term \(a_1 = 11\), the last term \(a_n = 99\), and the number of terms \(n = 9\). The sum \(S_n = \frac{n}{2}(a_1 + a_n) = \frac{9}{2}(11 + 99) = \frac{9}{2}(110) = 9 \times 55 = 495\). This confirms the sum calculated earlier.
  • The average of an arithmetic progression is simply the average of the first and last term: Average \(= \frac{a_1 + a_n}{2}\). In this case, Average \(= \frac{11 + 99}{2} = \frac{110}{2} = 55\). This provides a quicker way to find the average once the list of numbers is identified as an arithmetic progression.
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Important Questions from Average

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