Numbers

# The total number of months in a year are divisible by which numbers?

###### Wiki User

###### 2012-09-12 00:51:44

i dont even know

## Related Questions

###### Asked in Math and Arithmetic

### All numbers that are divisible by 6?

All numbers that are divisible by 6 are even numbers whose
digits add up to 3 or a multiple of 3.
Example : 284 is not wholly divisible by 6 as although it is
even, its digits total 2 + 8 + 4 = 14 which is not divisible by
3.
282 is divisible by 6 as it is even and its digits total 12
which is divisible by 3.

###### Asked in Math and Arithmetic

### How do you know if the number is divisible by 15?

All numbers divisible by 5 (of which 15 is a multiple) have a
final digit of 0 or 5.
All numbers divisible by 3 (of which 15 is a multiple) have the
sum of the digits totalling 3 or a multiple of 3.
Therefore, a number is divisible by 15 if the sum of its digits
total 3 or a multiple of 3 and its final digit is 0 or 5.
Example : 32085 ; 3 + 2 + 0 + 8 + 5 = 18 which is divisible by
3. Final digit 5.
This number is divisible by 15. (32085 ÷ 15 = 2139)
7420 : 7 + 4 + 2 + 0 = 13. This number is not divisible by
15.

###### Asked in Prime Numbers

### Show that there are no prime numbers from 200 to 210?

There are 2 prime numbers, they are 201 and 203
_____________________________
No.
201 is divisible by three: 3 x 67 = 201
203 is divisible by seven: 7 x 29 = 203
To show a number (n) is prime, you must show it is not divisible
by any prime up to the square root of n.
Any even number is divisible by two; any number ending in 5 is
divisible by five. So you only have to check the numbers ending in
1, 3, 7, or 9.
201 and 207 are divisible by three. Any easy check is to add the
digits. If that total is divisible by three, so is the original
number. That rules out 201 and 207.
203 is divisible by seven. Any easy (but not well known) test
for divisibility by seven is this. Take the ones digit, double it,
subtract from the remaining digits. In this case, for 203, take 3,
double it to 6, and subtract from 20 to get 14. 14 is divisible by
seven, so the original 203 is as well.
209 is divisible by eleven. A test for elevens is to add every
other digit to get one total, add the other digits to get a second
total, then take the difference of those totals. 2 + 9 = 11 for the
first total. 0 is the other total. 11 - 0 = 11 and that is
divisible by eleven.

###### Asked in Prime Numbers

### Is 176 a prime number?

No.
Go to www.ask.com and enter your question in the search box.
First result there is this which gives you the answer.
http://www.mathwarehouse.com/arithmetic/numbers/prime-number/prime-factorization.php?number=176
If the number ends in 0, 2, 4, 6, or 8, it is divisible by two.
If it ends in 0 or 5, it is divisible by five. If its digits total
some multiple of 3, it is divisible by 3.

###### Asked in Math and Arithmetic

### What are the numbers less than or equal to 50 that are divisible by 5?

If a number is at least 5, and ends in "5" or "0", then it is
divisible by 5. To further elaborate, If we are speaking of all
numbers, the answer is an infinite number. If we limit it to
positive integers, the answer is 50, 45, 40, 35, 30, 25, 20, 15,
10, 5, 0. Eleven total since 0 is a positive number.

###### Asked in Math and Arithmetic

### How many two digit numbers divisible by neither 3 nor 5?

48.
The first two digit number is 10, the last two digit number is
99, so there are 99 - 10 + 1 = 90 two digit
numbers
→ 10 ÷ 3 = 31/3 → first two digit number divisible by 3 is
4 x 3 = 12
→ 99 ÷ 3 = 33 → last two digit number divisible by 3 is
33 x 3 = 99
→ 33 - 4 + 1 = 30 two digit numbers divisible
by 3
→ 10 ÷ 5 = 2 → first two digit number divisible by 5 is
2 x 5 = 10
→ 99 ÷ 5 = 194/5 → last two digit number divisible by 5 is
19 x 5 = 95
→ 19 - 2 + 1 = 18 two digit numbers divisible
by 5
→ 30 + 18 = 48 two digit numbers divisible by 3
or 5 OR BOTH.
The numbers divisible by both are multiples of their lowest
common multiple: lcm(3, 5) = 15, and have been counted twice, so
need to be subtracted from the total
→ 10 ÷ 15 = 010/15 → first two digit number divisible by 15 is
1 x 15 = 15
→ 99 ÷ 15 = 69/15 → last two digit number divisible by 15 is
6 x 15 = 90
→ 6 - 1 + 1 = 6 two digit numbers divisible by
15 (the lcm of 3 and 5)
→ 48 - 6 = 42 two digit numbers divisible by 3
or 5.
→ 90 - 42 = 48 two digit numbers divisible by
neither 3 nor 5.

###### Asked in Math and Arithmetic

### What is the rule to determine if a number is divisible by 12?

That the number is divisible by 4* and the sum of its digits is
a multiple of 3.
*If the number has three of more digits then it is only
necessary to look at the tens and units to determine if it is
divisible by 4, as 4 is a factor of 100 and therefore of any
multiple of 100.
Examples : 75 : is not divisible by 4 although its digits total
12 which is a multiple of 3.
132 : is divisible by 4 as 32 is divisible by 4, and its digits
total 6 which is divisible by 3, then 132 is divisible by 12.