The three-digit Prime number that meets the given criteria is 373. This number is prime itself, with all its digits (3) being prime. Additionally, when considering the first two digits (37) and the last two digits (73) separately, both combinations also form prime numbers.
Not necessarily. Consider 444. The digits are not different. The first and second digits are not multiples of 3 The first digit is not greater than the second digit. In spite of all that, 444 is a 3-digit number
Oh, dude, you want to know the unit's digits of the product of the first 21 prime numbers? Well, let me casually tell you that the unit's digit of a product depends on the unit's digits of the numbers being multiplied. Since the unit's digit of all prime numbers greater than 5 is either 1, 3, 7, or 9, the product of the first 21 prime numbers will end in a unit's digit that is a result of multiplying these digits together. Cool, right?
The first squared number that is a multiple of 10 is three digits.
first A prime number (or a prime) is a natural number greater than 1 that has no positive divisors other than 1 and itself so 1 isnt a prime primes are 2 3 5 7 11 13 17 etc now by repeating digits if you mean the 1's in 11 then stop at 7 if you mean repeating a digit as in the tens spot of 11 and 13 stop at 11 and 3 and 13 are repating the 3 location....etc
All digits between the first non-zero digit and the last non-zero digits are significant. Some would argue that trailing 0s are significant since they are an indication of the precision of the number.
935
Not necessarily. Consider 444. The digits are not different. The first and second digits are not multiples of 3 The first digit is not greater than the second digit. In spite of all that, 444 is a 3-digit number
Oh, dude, you want to know the unit's digits of the product of the first 21 prime numbers? Well, let me casually tell you that the unit's digit of a product depends on the unit's digits of the numbers being multiplied. Since the unit's digit of all prime numbers greater than 5 is either 1, 3, 7, or 9, the product of the first 21 prime numbers will end in a unit's digit that is a result of multiplying these digits together. Cool, right?
The first digit would have to be 1, the remaining digits, zero.
5. Count the number of digits from the first non-zero digit to the last non-zero digit.
Normally a 2-digit number refers to an integer with two digits, the first of which is not 0. So the answer would be NO> But it is a number with 2 significant digits.
1348.
27
2.3
There are 10 digits, but for a three digit number the first number cannot be a 0. Thus: there is a choice of 9 digits for the first (and last digit which must be the same), with 10 choices of digit for the second (middle) digit, making 9 × 10 = 90 such palindromic numbers.
the first 35 digits of pi is... 3.14159265358979323846264338327950288
To form a two-digit number using the digits 0-9 without repetition, the first digit (the tens place) can be any digit from 1 to 9 (9 options), since a two-digit number cannot start with 0. The second digit (the units place) can then be any of the remaining 9 digits (including 0 but excluding the first digit). Therefore, the total number of two-digit numbers that can be formed is 9 (choices for the first digit) multiplied by 9 (choices for the second digit), resulting in 81 possible two-digit numbers.