[administration]

Q. a and b are natural numbers such that a > b > 1. If 8! is divisible by a2 × b2, then how many such sets (a, b) are possible?
1. 5
2. 6
3. 7
4. 8
5. null

[Wonder Boy]

8!=2⁷x3²x5x7

Now for 8! to be divisible by a²b² , factors which are useful is 2⁶ , 3² .

so now we have to make pairs from 2², 2², 2², 3² 

so the pairs will be [2²x2²](since a>b), [2²x2⁴], [2²x3²], [2⁴x3²], [2⁸x3²], [2²x(2×3)²], [2²x2²x(2×3)²], [2²x(2x2x3)²]

So total 7 sets are possible .

[Author]

Posts