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∙ 13y ago17
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∙ 13y ago73 is the largest 2 digit number that is both prime and has prime numbers for both of its digits.
73 is the largest two-digit number that is prime and has prime numbers for both of its digits.
There are only two smaller 3-digit numbers and both of them have repeated digits.
There are three such numbers: 12, 24 and 36.
Six three-digit numbers contain only the digits 5 and 6. This is assuming you mean both the digits of 5 and 6. If not than 8, as it would include 555 and 666. 556, 565, 566, 655, 656, 665
Since both of those numbers contains four digits, there are no three-digit numbers between them.
73 is the largest 2 digit number that is both prime and has prime numbers for both of its digits.
73 is the largest two-digit number that is prime and has prime numbers for both of its digits.
There are only two smaller 3-digit numbers and both of them have repeated digits.
There are three such numbers: 12, 24 and 36.
Six three-digit numbers contain only the digits 5 and 6. This is assuming you mean both the digits of 5 and 6. If not than 8, as it would include 555 and 666. 556, 565, 566, 655, 656, 665
There are 2000 possible five digit numbers that can be formed from the digits 02345 that are divisible by 2 or 5 or both. To be divisible by 2, the last digit must be even, namely 0, 2 or 4 (in the digits allowed). To be divisible by 5, the last digit must be 0 or 5. Thus to be divisible by 2 or 5 or both, the last digit must be 0, 2, 4 or 5 (a choice of 4). Presuming that a 5 digit number must be at least 10000, then: For the first digit there is a choice of 4 digits (2345); for each of these there is a choice of 5 digits (02345) for the second, making a total so far of 4 x 5 numbers; for each of these choices for the first and second digits there is a choice of 5 digits (02345) for the third digit making the total so far (4 x 5) x 5 numbers; for each of these choices for the first three digits there is a choice of 5 digits (02345) for the fourth digit making the total so far (4 x 5 x 5) x 5 numbers; for each of these choices for the first four digits there is a choice of 4 digits (0245 - as discussed above) for the last digit, giving a total of (4 x 5 x 5 x 5) x 4 numbers. So the total number of five digit numbers so formed is: number = 4 x 5 x 5 x 5 x 4 = 2000.
None. 1221 and 3443 are both 4-digit palindromes but no digit has remained the same between the two. First and fourth, second and third.
Yes. To find it, evaluate both irrationals until the numbers show a difference in one of their later digits. Truncate the irrationals after this digit, sum them, then divide by 2. Job done.
2
2
12 and 2412 and 2412 and 2412 and 24