Finding the Smallest Number to Add for Divisibility by 11

The question asks for the smallest number that should be added to 756896 so that the resulting number is a multiple of 11. To find this, we first need to understand the divisibility rule for 11 and determine the current remainder when 756896 is divided by 11.

Understanding Divisibility by 11

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

Applying the Divisibility Rule to 756896

Let's apply this rule to the number 756896. We identify the digits at odd and even positions starting from the rightmost digit.

Now, we calculate the sum of the digits at odd places and the sum of the digits at even places.

• Sum of digits at odd places (1st, 3rd, 5th): \(6 + 8 + 5 = 19\)
• Sum of digits at even places (2nd, 4th, 6th): \(9 + 6 + 7 = 22\)

Next, we find the difference between these two sums:

Difference = Sum of even place digits - Sum of odd place digits

Difference = \(22 - 19 = 3\)

Relating the Difference to the Remainder

The difference calculated using the divisibility rule (\(3\) in this case) is related to the remainder when the number is divided by 11. If the difference is positive, it is the remainder. If the difference is negative, we add 11 (or a multiple of 11) to get the positive remainder. If the difference is 0, the remainder is 0.

Here, the difference is 3. This means that when 756896 is divided by 11, the remainder is 3.

We can write this as: \(756896 = 11 \times q + 3\) (where \(q\) is the quotient).

Finding the Smallest Number to Add

We want to find the smallest number, let's call it \(x\), such that \(756896 + x\) is a multiple of 11.

Substituting the expression with the remainder:

\((11 \times q + 3) + x\) must be a multiple of 11.

For this expression to be a multiple of 11, the term \((3 + x)\) must be a multiple of 11.

We are looking for the smallest non-negative value for \(x\). The smallest positive multiple of 11 is 11. Therefore, we set \(3 + x\) equal to the smallest multiple of 11 that is greater than or equal to 3.

\(3 + x = 11\)

Solving for \(x\):

\(x = 11 - 3\)

\(x = 8\)

Wait! Let's re-check the remainder. The divisibility rule difference is often taken as the remainder mod 11, which can be negative. The remainder must be non-negative (0 to 10 when dividing by 11). The difference was \(19 - 22 = -3\). The remainder of \(-3\) when divided by 11 is \(8\) (since \(-3 + 11 = 8\)).

So the remainder is 8. This means \(756896 = 11 \times q + 8\).

We want \((11 \times q + 8) + x\) to be a multiple of 11. This requires \(8 + x\) to be a multiple of 11.

The smallest positive multiple of 11 is 11. The smallest value for \(8 + x\) that is a multiple of 11 and is \(\ge 8\) is 11.

\(8 + x = 11\)

Solving for \(x\):

\(x = 11 - 8\)

\(x = 3\)

So, the smallest number to be added is 3.

Verification

Let's add 3 to 756896:

\(756896 + 3 = 756899\)

Now, let's check if 756899 is divisible by 11 using the divisibility rule:

• Digits at odd places (1st, 3rd, 5th): 9, 8, 5. Sum = \(9 + 8 + 5 = 22\)
• Digits at even places (2nd, 4th, 6th): 9, 6, 7. Sum = \(9 + 6 + 7 = 22\)

Difference = Sum of even place digits - Sum of odd place digits = \(22 - 22 = 0\).

Since the difference is 0, 756899 is divisible by 11. Thus, adding 3 makes the number a multiple of 11, and 3 is the smallest positive number to achieve this.

Revision Table: Key Concepts for Divisibility by 11

Additional Information: Number Properties and Multiples

Understanding divisibility rules and number properties is fundamental in mathematics. Divisibility rules are shortcuts to determine if one number can be exactly divided by another without performing long division. Being a "multiple of 11" means the number can be expressed as \(11 \times k\) for some integer \(k\).

The concept of remainders is key here. Any integer \(N\) divided by 11 can be written in the form \(N = 11 \times q + r\), where \(q\) is the quotient and \(r\) is the remainder, with \(0 \le r < 11\). If \(r = 0\), the number is divisible by 11. If \(r \ne 0\), we need to add \(11 - r\) to \(N\) to get the next multiple of 11.

\(N + (11 - r) = (11 \times q + r) + (11 - r) = 11 \times q + 11 = 11 \times (q + 1)\)

This shows that adding \(11-r\) results in a multiple of 11. Since we want the smallest number to add, and the remainder \(r\) is between 0 and 10, \(11-r\) will give the smallest positive value needed (unless \(r=0\), in which case 0 is added).