The least number of numbers to be deleted from the set {1, 2, 3, ..., 14, 15} so the product is a perfect?

Question

square
a)1
b)2
c)3
d)4

Answer 1

let us list down all the prime factors of the numbers....

1: 1
2: 2
3: 3
4: 2, 2
5: 5
6: 2, 3
7: 7
8: 2, 2, 2
9: 3, 3
10: 2,5
11: 11
12: 2, 2, 3
13: 13
14: 2, 7
15: 3, 5

the product would have all these factors... lets strike out doubled factors from the list (the factors what would multiply to result in a perfect square; such as the two 2's in 12, 7 in 7 & 14 etc)... then we have the following remaining:

1:
2:
3:
4:
5: 5
6:
7:
8: 2
9:
10:
11: 11
12:
13: 13
14:
15:

the left out factors are: 2x5x11x13 or 10x11x13...so that the least number of numbers to be deleted is 3 (namely, 10, 11 & 13)

note:i came across this problem recently in an olympiad test

Answer 2

(c). Delete 10, 11 and 13. The product of the rest is 914457600=30240^2.

To see this, write out the prime factorizations of all the numbers. You have 11 2's, 6 3's, 3 5's, 2 7's, an 11 and a 13. Delete factors until you have an even number of all of them. That means getting rid of the 11, the 13, a 5 and a 2.

Answer 3

Resolve each of the numbers into its prime factors
and count how many times each appears in the product.
If it appears an odd number of times knock out one
of its factors. If it appears an even number of times
leave it alone.
Unless I miscounted I got 11 2's
6 3's
3 5's
2 7's
1 11
1 13
So we must knock out one 2, one 5, one 11 and one 13
or 4 factors to have all exponents even, i.e., a
square.

Answer 4

Wow, is there an approach to this problem other than hunt and peck?

n*(n-1)*(n-2)*...*1 = k^2 ?

Write out the prime factorization:
2^7*3^5 etc

need all exponents to be even

Answer 5

4 numbers, 5,8,11 and 13