Find the sum of all positive integers less than 1000 ending in 3 or 7.
I see it as
(3 + 13 + 23 + + 993) + (7 + 17 + ... + 997)
two arithmetic series,
there are 100 terms in each series with a= 3 and d = 10 and a=7 and d=10
using :
sum= (n/2)(first + last)
sum = (100/2)(3+993) + (100/2)(7+997)
= 100,000
First, you need to list all of the integers that match the condition. Do you know what these integers are?
Positive integers less than 1000 consist of {1 ... 999}.
This may also be written {001 ... 999}.
So you need to count, {003, 007, 013, ... 987, 993, 997}.
Which is the same result as counting {00, 01, ... 98, 99} twice over. (Once for ??3, once for ??7).
Count the number of integers from 0 and 99 inclusive. Double that.
Thanks, guys!
No cheating >:(
To find the sum of all positive integers less than 1000 ending in 3 or 7, we can use the concept of an arithmetic series.
First, let's find the total number of terms that end in 3 or 7.
We observe that there is one number ending in 3 (3) and one number ending in 7 (7) in the range of 1 to 10. The same pattern repeats for every multiple of 10. So, for every 10 numbers, we have 2 numbers ending in 3 or 7.
To find the number of multiples of 10 in the range of 1 to 1000, we can divide 1000 by 10: 1000 / 10 = 100.
Thus, there are 100 sets of 10 numbers within the range.
Since each set has 2 numbers ending in 3 or 7, we can multiply 2 by 100 to get the total number of terms ending in 3 or 7: 2 * 100 = 200.
Now we need to find the sum of these 200 terms.
To do this, we can use the formula for the sum of an arithmetic series:
Sum = (n/2)(first term + last term)
The first term is 3, and the last term is 997 (the largest number less than 1000 ending in 3 or 7).
Using the formula, the sum of the 200 terms ending in 3 or 7 can be calculated as:
Sum = (200/2)(3 + 997) = (100)(1000) = 100,000.
Therefore, the sum of all positive integers less than 1000 ending in 3 or 7 is 100,000.