3 digit number divisible by 3 and two digits repeated.....???

DearShubham,

For the number to be divisible by 4, the number formed by the last two digits must be divisible by 4. With the digits 1 through 6, you can use the 3 last digits 2, 4 and 6, and each of these can be combined with half of the digits to form a 2-digit number divisible by 4, so there are 9combinations for the last two digits. The other two digits can be chosen arbitrarily, so there are 36combinations for those, for a total of 9\cdot 36=324 different numbers

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Aman Bansal

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Hey

suppose 114

to test this = 1+1+4 =6 .

six is divisable by 3 so

the given number also divisable by 3.

there are 66 possiblities are there for your question.

they are = {114,117,225,228,330,336,339,441,448,552,558,660,669,771,774,882,885,990,993,996} there is also possibloe of repetitions  other are 300,600,900.