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Q: Is 195 divisible by 5

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195 is composite because it is divisible by 5.

1, 3, 5, 13, 15, 39, 65, 195.

All multiples of 10 end in a zero To be divisible by 10, the last digit must by a 0 The last digit of 195 is a 5 which is NOT a 0, so 195 is NOT divisible by 10.

Multiples of 195.

Because any number ending with 5 or 0 is divisible by 5 except of course 0 by itself

No - 195/2 = 97.5

Yes. 195/3=65. 65x3=195.

You can find a common denominator by multiplying together the denominators of your fractions. For example, try 13 * 15 = 195 (you don't need to use the 5 because anything divisible by 15 will also be divisible by 5). You know that 195 is divisible by 13 and 15 (because those are the two numbers you multiplied together to get it), and by 5 because 195 ends in a 5. Now convert the numerators so you have equivalent fractions: 5/13 = 75/195 2/5 = 78/195 7/15 = 91/195 So 7/15 is the greatest. But sometimes it's easier just to convert your fractions to decimals by dividing the numerators by the denominators: 5/13 = .38461538461538 2/5 = .4 7/15 = .46666666666666 (repeating forever)

if all the digits add up to a number divisible by 3 like 195 195= 1+9+5= 15...15/3=5 so it is divisible by 3

135. Every increase of 30 is also. Such as 165, 195, 225.

585 is divisible by these numbers: 1 3 5 9 13 15 39 45 65 117 195 and 585.

Any of its multiples

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