How Many Two Digit Positive Integers Exist That Leave Remainder 5 When Divided By 7 And What Is The Sum Of All These Integers.
How Many Two Digit Positive Integers Exist That Leave Remainder 5 When Divided by 7 and What is the Sum of All These Integers?
In this article, we will explore the concept of remainders when dividing integers by a certain number. Specifically, we will focus on finding the number of two-digit positive integers that leave a remainder of 5 when divided by 7 and then calculate the sum of all these integers.
When an integer is divided by another integer, the remainder is the amount left over after the division. For example, if we divide 17 by 7, we get a quotient of 2 and a remainder of 3. This can be represented as:
17 = 7(2) + 3
In this case, the remainder is 3.
To find the two-digit positive integers that leave a remainder of 5 when divided by 7, we can start by listing the first few multiples of 7 and then add 5 to each multiple to get the desired integers.
7(1) + 5 = 12 7(2) + 5 = 19 7(3) + 5 = 26 7(4) + 5 = 33 7(5) + 5 = 40 7(6) + 5 = 47 7(7) + 5 = 54 7(8) + 5 = 61 7(9) + 5 = 68 7(10) + 5 = 75 7(11) + 5 = 82 7(12) + 5 = 89 7(13) + 5 = 96
We can see that the first 12 two-digit positive integers that leave a remainder of 5 when divided by 7 are 12, 19, 26, 33, 40, 47, 54, 61, 68, 75, 82, and 89.
To calculate the sum of the integers, we can use the formula for the sum of an arithmetic series:
Sum = (n/2)(a + l)
where n is the number of terms, a is the first term, and l is the last term.
In this case, the first term is 12, the last term is 96, and the number of terms is 12.
Sum = (12/2)(12 + 96) Sum = 6(108) Sum = 648
Therefore, the sum of all the two-digit positive integers that leave a remainder of 5 when divided by 7 is 648.
In conclusion, we have found that there are 12 two-digit positive integers that leave a remainder of 5 when divided by 7. These integers are 12, 19, 26, 33, 40, 47, 54, 61, 68, 75, 82, and 89. We have also calculated the sum of these integers, which is 648.
The final answer is 12 two-digit positive integers that leave a remainder of 5 when divided by 7, and the sum of all these integers is 648.
Frequently Asked Questions (FAQs) About Two Digit Positive Integers Leaving a Remainder of 5 When Divided by 7
A: The smallest two-digit positive integer that leaves a remainder of 5 when divided by 7 is 12. This is because 7(1) + 5 = 12.
A: The largest two-digit positive integer that leaves a remainder of 5 when divided by 7 is 96. This is because 7(13) + 5 = 96.
A: There are 12 two-digit positive integers that leave a remainder of 5 when divided by 7. These integers are 12, 19, 26, 33, 40, 47, 54, 61, 68, 75, 82, and 89.
A: The sum of all the two-digit positive integers that leave a remainder of 5 when divided by 7 is 648. This can be calculated using the formula for the sum of an arithmetic series: Sum = (n/2)(a + l), where n is the number of terms, a is the first term, and l is the last term.
A: Yes, the formula to find the nth two-digit positive integer that leaves a remainder of 5 when divided by 7 is:
an = 7n + 5
where an is the nth two-digit positive integer that leaves a remainder of 5 when divided by 7, and n is the position of the integer in the sequence.
A: Yes, the formula to find the sum of the first n two-digit positive integers that leave a remainder of 5 when divided by 7 is:
Sum = (n/2)(a + l)
where n is the number of terms, a is the first term, and l is the last term.
A: If you don't know the first term, you can start by listing the first few multiples of 7 and then add 5 to each multiple to get the desired integers. Alternatively, you can use the formula an = 7n + 5 to find the nth two-digit positive integer that leaves a remainder of 5 when divided by 7.
A: Yes, the list of all the two-digit positive integers that leave a remainder of 5 when divided by 7 is:
12, 19, 26, 33, 40, 47, 54, 61, 68, 75, 82, 89, 96
Note that this list only includes two-digit positive integers that leave a remainder of 5 when divided by 7.