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A number is divisible by 2^n if the last 'n' digits are divisible by n

DOES IT REALLY WORK??? Let's Check

1. 12 /2

Number above is divided by 2^1 so the last 1 digit will be divisible by 2 So 2/2 , Remainder = 0

2. 5896/8

Number above is divided by 2^3 so the last 3 digits will be divisible by 8 So 896/8 , Remainder = 0 YES IT DOES!!! Lets start Applying

1.12348/4

Last 2 digits of the number should be divisible by 4 because 2^2 = 4,

since it's 2^2 we involve last 2 digits So 48/4, Remainder = 0

2.27365/8

Last 3 digits of the number should be divisible by 8 because 2^3 = 8,

since it's 2^3 we involve last 3 digits So 365/8 Remainder = 5

3. 1246576543298/32

Last 5 digits of the number should be divisible by 32 because 2^5 = 32,

since it's 2^5 we involve last 5 digits

So 43298/32, Remainder = 2

4. (2243576)^2/16

Last 4 digits of the number should be divisible by 16 because 2^4 = 16,

since it's 2^4 we involve Last 4 digits So (3576)^2/16, Remainder = 8^2 = 64/8 = 0

5. 123456........989100/16

Last 4 digits of the number should be divisible by 16 because 2^4 = 32,

since it's 2^4 we involve

Last 4 digits So 9100/16, Remainder = 6

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