Hi Mehta17,I can see exactly where this slipped. You read "11 five-point baskets" as if it added up to
11 ×
5 =
55, but the question multiplies the
point values together, not adds them.
So let's actually build your combination as a
product:
-
11 baskets worth
5 points each -
5 ×
5 × ... ×
5 (eleven times) =
511-
1 basket worth
11 points - ×
11-
2 baskets worth
2 points each - ×
2 ×
2That product is
511 × 11 × 4, which is in the tens of millions - nowhere near
2,420. So this combination doesn't actually work.
Why the answer is forcedHere's the deeper reason it's unique:
2,420 breaks into exactly one set of primes.
2,420 =
22 ×
5 ×
112Every basket value (
2,
5,
11,
13) is either prime or made of these primes, so the point values
have to be exactly these factors: two
2's, one
5, and two
11's. There's no freedom - you can't slip in extra
5's, because there's only
one5 in
2,420. The
11 appears exactly
twice, so Irwin scored
two 11-point baskets. That's
C.
Quick check to lock it in: try to make
20 as a product of allowed basket values.
20 =
2 ×
2 ×
5. Could you instead use "two 5-point baskets"? That would need
5 ×
5 =
25 ≠
20. The prime factorization tells you the
only breakdown - there's no second option hiding.
Answer: CMehta17
Since there is no restriction in number of basket
what if he score 11*5 = 55 (11 baskets with 5 points) 1 with 11 points and 2 with 2 points
Still it would 2420