Q: How many 5 digits can be formed using the digits 0 2 3 4 5 when repetition is allowed that the number formed is divisible by 2 and 5 or both?

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If repetition of digits isn't allowed, then no13-digit sequencescan be formed from only 5 digits.

64 if repetition is allowed.24 if repetition is not allowed.

None. Nine digit number cannot be formed using only five digits 1,2,3,4,5 in the case none of the digits can repeat. -------- Egad! I misread the question, apologies to the questioner. And thanks to Miroslav.

125

64 if repetition is allowed, 6 otherwise.64 if repetition is allowed, 6 otherwise.64 if repetition is allowed, 6 otherwise.64 if repetition is allowed, 6 otherwise.

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If repetition of digits isn't allowed, then no13-digit sequencescan be formed from only 5 digits.

There are 2000 such numbers.

64 if repetition is allowed.24 if repetition is not allowed.

None. Nine digit number cannot be formed using only five digits 1,2,3,4,5 in the case none of the digits can repeat. -------- Egad! I misread the question, apologies to the questioner. And thanks to Miroslav.

125

64 if repetition is allowed, 6 otherwise.64 if repetition is allowed, 6 otherwise.64 if repetition is allowed, 6 otherwise.64 if repetition is allowed, 6 otherwise.

125 There are five choices for each of the three digits (since repetition is allowed). So there are 5*5*5=125 combinations.

Possible solutions - using your rules are:- 11,13,17,31,33,37,71,73 &77

It is 415968.

20 of them, if repetition is not allowed.

125 different 4 digits numbers. To be divisible by 4, the last two digits must be divisible by 4, which means they must be: 12, 24, 32, 44 or 52 - 5 possible choices. For each of these there are 5 choices for the first digit and 5 for the second, meaning: total = 5 x 5 x 5 = 125

Just six numbers... 345, 354, 435, 453, 534 & 543