1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 + 25 + 27 + 29 + 31 + 33 + 35 + 37 + 39 + 41 + 43 + 45 + 47 + 49 + 51 + 53 + 55 + 57 + 59 + 61 + 63 + 65 + 67 + 69 + 71 + 73 + 75 + 77 + 79 + 81 + 83 + 85 + 87 + 89 + 91 + 93 + 95 + 97 + 99 = 2500
There are 80 such integers.
The squares -- nine of them.
8 of them.
1, 4, 9, 16, 25, 36, 49, 64, 81, 100 -- the square numbers.
I'm assuming the question should read n greater than 10 and less than 100 and there are 8 numbers that satisfy this,1223344556677889So the answer is dhttp://www.webanswers.com/share-question.cfm?q=how-many-integers-n-greater-than-and-less-than-100-are-there-such-that-if-the-digits-of-n-are-the-is-1379ff| http://www.webanswers.com/answer/1566078/education/how-many-integers-n-greater-than-and-less-than-100-are-there-such-that-if-the-digits-of-n-are-the-is-1379ff| http://www.webanswers.com/report-abuse.cfm?q=how-many-integers-n-greater-than-and-less-than-100-are-there-such-that-if-the-digits-of-n-are-the-is-1379ff&p=1566078
The sum of all the digits of all the positive integers that are less than 100 is 4,950.
Sum of all such integers less than 100 would be 416 .
There are 80 such integers.
The squares -- nine of them.
32
All of them from 45 to 99 ... 55 integers.
8 of them.
81
There are 20.
The sum of such numbers is 504.
3, 6, 9... just continue adding 3 at a time.
To determine how many positive integers less than 100 have reciprocals with terminating decimal representation, we need to consider the prime factorization of each integer. Any positive integer whose prime factorization consists only of 2s and/or 5s will have a reciprocal with a terminating decimal representation. In the range of positive integers less than 100, there are 49 numbers that are powers of 2, 5, or their products (2^0, 2^1, 2^2, ..., 2^6, 5^0, 5^1, 5^2). Therefore, there are 49 positive integers less than 100 with reciprocals having terminating decimal representations.