What is the largest number 2 digit number with only 4 factors?

Answer 1

Well, isn't that a happy little question! The largest 2-digit number with only 4 factors is 77. You see, 77 can be divided evenly by 1, 7, 11, and 77 itself, making it a special number indeed. Just like every number, it has its own unique beauty and charm.

Answer 2

Oh, dude, that's an easy one. The largest 2-digit number with only 4 factors is 77. Yeah, it's like 1, 7, 11, and 77 itself. So, if you're ever at a party and someone asks for a fun fact, you can impress them with this little gem.

Answer 3

Well, honey, the largest 2-digit number with only 4 factors is 77. It's got 1, 7, 11, and 77 as its only factors. So, there you have it, a sassy little number with just enough factors to keep things interesting.

Answer 4

The largest 2-digit number with only 4 factors is 77. The factors of 77 are 1, 7, 11, and 77. In general, numbers with exactly 4 factors are of the form p^3, where p is a Prime number. In this case, 77 can be expressed as 7^3.

Answer 5

Short answer:Depending on whether or not you consider 1 and/or the number itself valid factors, it is either

99 with {3, 9, 11, 33}

or

81 with {1, 3, 9, 27} or {3, 9, 27, 81} respectively

or

95 with {1, 5, 19, 95}.

Long answer:

In general, if a number n has the prime factorization

n = p1k1 · p2k2· p3k3 · ... · pmkm

then the number of divisors including 1 and n itself is

#{z : z|n} = (k1+1) · (k2+1) · ... · (km+1)

because you can pick each prime factor not at all (0 times), 1 time, 2 times, ... or k times in order to compose a distinct divisor of n.

Thus, if you e.g. only ask for non-trivial divisors you need to find ki such that

(k1+1) · (k2+1) · ... · (km+1) -2 = 4.

In other words

(k1+1) · (k2+1) · ... · (km+1) = 6 = 2 · 3 · 1 · 1 · ... · 1,

which is only possible if one bracket yields 2, another 3 and all the remaining ones 1.

So k1=1, k2=2 and k3=k4=...km=0, thus n must consist of precisely two unique primes:

n = p11 · p22.

Indeed, then the non-trivial factors are: p1, p2, p1·p2 and p22.

Luckily, the highest two-digit number (being 99) already fulfills this condition (p1=11, p2=3). :-)

Good day

Answer 6

95