Question: Find the sum of all odd numbers greater than 100 and less than 200 that are multiples of 3, and subtract the sum of all even numbers greater than 100 and less than 200 that are multiples of 5.

To solve this problem, we need to find two sets of numbers. The first set comprises odd numbers greater than 100 and less than 200 that are divisible by 3. We find these numbers to be 105, 111, 117, 123, 129, 135, 141, 147, 153, 159, 165, 171, 177, 183, 189, and 195. The sum of these numbers is 2499. The second set comprises even numbers greater than 100 and less than 200 that are divisible by 5. These numbers are 110, 120, 130, 140, 150, 160, 170, 180, and 190. The sum of these numbers is 1370. The final step is to subtract the sum of even numbers from the sum of odd numbers: 2499 - 1370 = 1129. Therefore, the answer is 1129.

Math question: Calculate the sum of positive odd integers between 101 and 200 (inclusive) that are divisible by 3.

To find the sum of positive odd integers between 101 and 200 that are divisible by 3, we need to identify the values that satisfy both conditions. Out of the given range, the first odd integer divisible by 3 is 105, while the last one is 199. To calculate the sum, we can use the arithmetic series formula: Sum = (n/2)(first term + last term), where n is the number of terms. We can find the number of terms by using the formula for the nth term of an arithmetic series: an = a1 + (n-1)d, where an is the last term, a1 is the first term, and d is the common difference. Here, a1 = 105, an = 199, and d = 3. When we rearrange the formula to solve for n, we get: n = (an - a1 + d) / d. Plugging in the given values, we have n = (199 - 105 + 3) / 3 = 99/3 = 33. So, there are 33 positive odd integers between 101 and 200 that are divisible by 3. Now, let's find the sum. The first term is 105, and the last term is 199. Plugging these values into the sum formula: Sum = (33/2)(105 + 199) = 16.5(304) = 5016. Therefore, the sum of positive odd integers between 101 and 200 that are divisible by 3 is 5016.

Math question: What is the sum of all odd numbers from 101 to 200 (inclusive)?

To find the sum of odd numbers from 101 to 200, we can use the formula for the sum of an arithmetic series. The first odd number in the range is 101, and the last one is 199. There are (199 - 101) / 2 + 1 = 50 terms in total. The common difference is 2 since we are dealing with odd numbers. Using the formula for the sum of an arithmetic series, which is Sum = (number of terms / 2) * (first term + last term), we substitute the values: Sum = (50 / 2) * (101 + 199) = 25 * 300 = 7500. Therefore, the sum of odd numbers from 101 to 200 is 7500.

Max	9223372036854775807
25835	Positive integer
0b10001100	Binary number
01771452	Octal number
0x129abf	Hex number
5.012 * 10^7	Scientific notation
2^4 * 5	Factorized form
*	Random number

101 (number)

101 is an odd 3-digits prime number following 100 and preceding 102. In scientific notation, it is written as 1.01 × 10^2. The sum of its digits is 2. It has a total of 1 prime factors and 2 positive divisors. There are 100 positive integers (up to 101) that are relatively prime to 101.

Interesting facts

101 has 2 divisors, whose sum is 102
The reverse of 101 is 101
Previous prime number is 1

Basic properties

Is Prime? yes
Number parity odd
Number length 3
Sum of Digits 2
Digital Root 2

Name

Name	one hundred one

Notation

Scientific notation	1.01 × 10^2
Engineering notation	1.01 × 10^0

Prime Factorization of 101

Prime Factorization 101

Composite number

ω	Distinct Factors	1	Total number of distinct prime factors
Ω	Total Factors	1	Total number of prime factors
rad	Radical	101	Product of the distinct prime numbers

The prime factorization of 101 is 101. Since it has a total of 1 prime factors.

Most popular turbo-numbers

4, 9, 91, 35, 135, 750, 7717, 2222, 31995, 42167, 876695, 237851, 8321094, 6909183, 85684694, 94493273, 839689672, 709193585, 3013374712, 1504753955

Basic calculations (n = 101)

Multiplication

n×y

n×2	202
n×3	303
n×4	404
n×5	505

Division

n÷y

n÷2	50.5
n÷3	33.666666666667
n÷3	25.25
n÷5	20.2

Exponentiation

n^y

n²	10201
n³	1030301
n⁴	104060401
n⁵	10510100501

Nth Root

^y√n

²√n	10.049875621121
³√n	4.6570095078038
⁴√n	3.1701538797227
⁵√n	2.5168902290141

Divisors of 101

1, 101

2 divisors

τ	Total Divisors	2	Total number of the positive divisors of n
σ	Sum of Divisors	102	Sum of all the positive divisors of n
s	Aliquot Sum	1	Sum of the proper positive divisors of n
A	Arithmetic Mean	51	Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n))
G	Geometric Mean	10.099504938362	Returns the nth root of the product of n divisors
H	Harmonic Mean	1.9803921568627	Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors

The number 101 can be divided by 2 positive divisors. The sum of these divisors (counting 101) is 102, the average is 51.

Base conversion 101

Base	System	Value
2	Binary	1100101
3	Ternary	10202
4	Quaternary	1211
5	Quinary	401
6	Senary	245
8	Octal	145
10	Decimal	101
12	Duodecimal	85
16	Hexadecimal	65
20	Vigesimal	51
36	Base36	2t

Cryptographic Hash Functions

md5	38b3eff8baf56627478ec76a704e9b52
sha1	dbc0f004854457f59fb16ab863a3a1722cef553f
sha256	16dc368a89b428b2485484313ba67a3912ca03f2b2b42429174a4f8b3dc84e44
sha512	be37ccebe21815559666b60338ec1492670b8fd2bf6cc63c5c943639ddcc50981003846b75b9e97ad0c0c19484292f59b1d30b45c7b07d1f8973bae68a3b8431
ripemd-160	fe9362e30831f55eeb1f0d837cd5f98848386ef9

101 as geometric shapes

Circle

Radius = n

Diameter	202
Circumference	634.60171602514
Area	32047.386659269

Sphere

Radius = n

Volume	4315714.7367816
Surface area	128189.54663708
Circumference	634.60171602514

Square

Length = n

Perimeter	404
Area	10201
Diagonal	142.83556979968

Cube

Length = n

Surface area	61206
Volume	1030301
Space diagonal	174.93713156446

Equilateral Triangle

Length = n

Perimeter	303
Area	4417.1625720025
Altitude	87.468565782228

Triangular Pyramid

Length = n

Surface area	4417.1625720025
Volume	121422.13729388
Height	47.611856599894