Q: What three digit number has the most factors?

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If you're asking about distinct prime factors, there are eight numbers tied with three of them. If not, 64 has six twos.

60, 72, 84, 90 and 96 each have twelve factors.

The squares of most 2 digit prime numbers. For example 2809, which is the square of 53, has 3 factors : 1, 53 and 2809 itself.

There is NO number with the most number of factors.

One way to solve this is to multiply the lowest prime numbers together, starting from 2: 2x3=6. 6x5=30. Multiplying by 7 would yield a 3-digit number, so we are looking for all 2-digit numbers with 3 distinct prime factors. We already have 30. 2x3x7=42. 2x3x11=66. 2x3x13=78. 2x5x7=70. So, to answer the original question, the 2-digit numbers that have the most factors in their prime factorization are, in ascending order, 30, 42, 66, 70, and 78.

Related questions

512 has 9 prime factors

8192 has 13 prime factors

If you're asking about distinct prime factors, there are eight numbers tied with three of them. If not, 64 has six twos.

60, 72, 84, 90 and 96 each have twelve factors.

as usually in most three digit left

60, 72, 84, 90 and 96 each have twelve factors.

Actually the most common winning three digit numbers in the Thailand lotto are 231, 453 and 234

The squares of most 2 digit prime numbers. For example 2809, which is the square of 53, has 3 factors : 1, 53 and 2809 itself.

If you're asking about distinct prime factors, there are eight numbers tied with three of them. If not, 64 has six twos.

There cannot be a proper three digit number between 10 and 20. It must be a two digit number. If you consider a 2-digit number between 10 and 20, then its tens digit MUST be 1 and so MUST be odd. So most of this question is either wrong or irrelevant. Are you sure you have not cut and pasted bits from two questions?

For a 3 digit number, the left most or the most significant digit cannot be zero. So it can be 1,2,3,4,5,6,7,8 or 9 which is 9 possibilities. The middle number can be 0,1,2,3,4,5,6,7,8,9 which is 10 possibilties but one of the digits has been chosen already as the first digit, so the possibilities are only 9. The right most number can be 0,1,2,3,4,5,6,7,8,9 which is 10 possibilities but two of the digits have been already used by the left most and the middle digits. That leaves only 8 possibilities. So the total number of three digit numbers that have three distinct digits is 9 x 9 x 8 = 81 x 8 = 648 possibilities

For this type of "the largest number which..." questions, you need to advance from left to right, using the largest possible digit in each case. For the first two digits, that would be 9, for the third digit (the right-most digit), the largest digit which will make this possible is an 8.