Math question: If I multiply a number by 3, add 20, and then subtract 5, the result is 155. What is the original number?

To solve the problem, let's assume the unknown number is 'x'. According to the given information, we can write the equation as 3x + 20 - 5 = 155. Simplifying this equation leads to 3x + 15 = 155. Next, we subtract 15 from both sides, which gives 3x = 140. To solve for 'x', we divide both sides by 3, resulting in x = 140/3. Thus, the original number is approximately 46.67.

Question: What is the sum of all odd numbers from 1 to 155?

To find the sum of all odd numbers from 1 to 155, we can use the formula for the sum of an arithmetic series. First, we need to find the number of terms in the series. The last odd number is 155, and the first odd number is 1. So, the number of terms is (155 - 1)/2 + 1 = 78. Next, we calculate the sum using the formula S = (n/2)(a + l), where S is the sum, n is the number of terms, a is the first term, and l is the last term. Plugging in the values we have, S = (78/2)(1 + 155) = 39 * 156 = 6084. Therefore, the sum of all odd numbers from 1 to 155 is 6084.

Math Question: Consider a number whose hundreds digit is 1 and units digit is 5, which ranges from 100 to 199. How many such numbers exist?

To determine the total count of such numbers, we have to consider the tens digit. From 10 to 19, there is only one possibility (15). From 110 to 119, again only one number satisfies the condition (115). Similarly, from 210 to 219, 310 to 319, and so on, we observe that exactly one number satisfies the condition. Thus, for each hundred, we have exactly one number meeting the criteria. There are 100 possible hundreds digits (ranging from 1 to 100) and one corresponding number for each hundred. Therefore, the answer is 100.

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

155 (number)

155 is an odd 3-digits composite number following 154 and preceding 156. In scientific notation, it is written as 1.55 × 10^2. The sum of its digits is 11. It has a total of 2 prime factors and 4 positive divisors. There are 120 positive integers (up to 155) that are relatively prime to 155.

Interesting facts

155 has 4 divisors, whose sum is 192
The reverse of 155 is 551
Previous prime number is 31

Basic properties

Is Prime? no
Number parity odd
Number length 3
Sum of Digits 11
Digital Root 2

Name

Name	one hundred fifty-five

Notation

Scientific notation	1.55 × 10^2
Engineering notation	1.55 × 10^0

Prime Factorization of 155

Prime Factorization 31 × 5

Composite number

ω	Distinct Factors	2	Total number of distinct prime factors
Ω	Total Factors	2	Total number of prime factors
rad	Radical	155	Product of the distinct prime numbers

The prime factorization of 155 is 31 × 5. Since it has a total of 2 prime factors.

Most popular turbo-numbers

8, 7, 56, 78, 834, 457, 9337, 5761, 73392, 10177, 904320, 463175, 6775231, 462269, 47645885, 61261763, 999898499, 485946794, 6629714702, 8017972789

Basic calculations (n = 155)

Multiplication

n×y

n×2	310
n×3	465
n×4	620
n×5	775

Division

n÷y

n÷2	77.5
n÷3	51.666666666667
n÷3	38.75
n÷5	31

Exponentiation

n^y

n²	24025
n³	3723875
n⁴	577200625
n⁵	89466096875

Nth Root

^y√n

²√n	12.449899597989
³√n	5.3716853549448
⁴√n	3.5284415253747
⁵√n	2.7419929866413

Divisors of 155

1, 5, 31, 155

4 divisors

τ	Total Divisors	4	Total number of the positive divisors of n
σ	Sum of Divisors	1	Sum of all the positive divisors of n
s	Aliquot Sum	-154	Sum of the proper positive divisors of n
A	Arithmetic Mean	0.25	Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n))
G	Geometric Mean	1	Returns the nth root of the product of n divisors
H	Harmonic Mean	4	Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors

The number 155 can be divided by 4 positive divisors. The sum of these divisors (counting 155) is 1, the average is 0.25.

Base conversion 155

Base	System	Value
2	Binary	10011011
3	Ternary	12202
4	Quaternary	2123
5	Quinary	1110
6	Senary	415
8	Octal	233
10	Decimal	155
12	Duodecimal	10b
16	Hexadecimal	9b
20	Vigesimal	7f
36	Base36	4b

Cryptographic Hash Functions

md5	2a79ea27c279e471f4d180b08d62b00a
sha1	9d8974baddfc0e53300829f37e5fc88b0f5ce61b
sha256	210e3b160c355818509425b9d9e9fd3ea2e287f2c43a13e5be8817140db0b9e6
sha512	00aae04a2ba47c629332328c83d8a90ef443da6199bd76fa3b188b8654de8376e469c44f6986dbf7853d2e532efb04127dba65d072a4bb3aea6c4129044f5b00
ripemd-160	175ab53ec2382b2ce3c2d2147bc97fb18dc7be9a

155 as geometric shapes

Circle

Radius = n

Diameter	310
Circumference	973.89372261284
Area	75476.763502495

Sphere

Radius = n

Volume	15598531.123849
Surface area	301907.05400998
Circumference	973.89372261284

Square

Length = n

Perimeter	620
Area	24025
Diagonal	219.20310216783

Cube

Length = n

Surface area	144150
Volume	3723875
Space diagonal	268.46787517318

Equilateral Triangle

Length = n

Perimeter	465
Area	10403.130162961
Altitude	134.23393758659

Triangular Pyramid

Length = n

Surface area	10403.130162961
Volume	438862.87746518
Height	73.06770072261