Question

The sum of all two digit numbers that give a remainder 2 when they are divided by 7 is ______

• A. 552
• B. 654
• C. 658
• D. 684

Hint:

The numbers are of the form 7k+2 between 10 and 99, forming an AP from 16 to 93; use the AP sum formula.

Answer 1

The Correct Option is B

To find the sum of all two-digit numbers that give a remainder of 2 when divided by 7, we need to first identify the relevant two-digit numbers and then add them up. Here's a step-by-step explanation:

1. Identify the sequence: 

We need to find all two-digit numbers of the form \(7k + 2\) where \(k\) is an integer such that the result is a two-digit number.

1. Determine the range:

The smallest two-digit number is 10. So, we start from \(7k + 2 \geq 10\). Solving for \(k\): \(7k + 2 \geq 10 \Rightarrow 7k \geq 8 \Rightarrow k \geq \frac{8}{7}\). The smallest integer value for \(k\) is 2.

The largest two-digit number is 99. So, we have \(7k + 2 \leq 99\). Solving for \(k\): \(7k + 2 \leq 99 \Rightarrow 7k \leq 97 \Rightarrow k \leq \frac{97}{7}\). The largest integer value for \(k\) is 13.

1. List the sequence:

The two-digit numbers that satisfy the given condition are: \(7 \times 2 + 2 = 16\), \(7 \times 3 + 2 = 23\), \(7 \times 4 + 2 = 30\), ..., up to \(7 \times 13 + 2 = 93\).

Thus, the sequence is: 16, 23, 30, 37, 44, 51, 58, 65, 72, 79, 86, 93.

1. Calculate the sum:

The sequence is an arithmetic sequence where: \(a = 16\), \(l = 93\) (last term), and the common difference \(d = 7\).

The number of terms \(n\) is given by: \(n = \frac{l - a}{d} + 1 = \frac{93 - 16}{7} + 1 = 12\).

The sum \(S_n\) of the first \(n\) terms of an arithmetic sequence is: \(S_n = \frac{n}{2} \times (a + l)\). Substituting the values: \(S_{12} = \frac{12}{2} \times (16 + 93) = 6 \times 109 = 654\).

Thus, the sum of all two-digit numbers that give a remainder 2 when divided by 7 is 654.

Answer 2

Step 1: Write the numbers in the right form.
A number that leaves remainder 2 when divided by 7 looks like \(7k + 2\) for some whole number \(k\).

Step 2: Find the smallest and largest two-digit values.
For \(k = 2\), \(7(2) + 2 = 16\), the smallest two-digit number of this form. For \(k = 13\), \(7(13) + 2 = 93\), the largest two-digit number of this form (since \(k = 14\) gives 100, a three-digit number).

Step 3: List the numbers and find how many there are.
The numbers are 16, 23, 30, ..., 93, which form an arithmetic progression (AP) with common difference 7. Number of terms \(n = \frac{93 - 16}{7} + 1 = \frac{77}{7} + 1 = 11 + 1 = 12\).

Step 4: Add the AP using the sum formula.
Sum of an AP with \(n\) terms, first term \(a\), last term \(l\), is \(\frac{n}{2}(a + l)\). Here \(n = 12\), \(a = 16\), \(l = 93\), so sum \(= \frac{12}{2}(16 + 93) = 6 \times 109 = 654\).

Final Answer:
The sum of all such two-digit numbers is 654. \[ \boxed{654} \]