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Arranging the prime factors we can clearly see here that 11^2 is a factor.
hence @
smash C
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2420= (2^2) x (5^1) x (11^2)

So, Irwin can score 2 11-point baskets.
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The answer is 2 (11*11*20)
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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
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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 × 2

That 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 forced

Here's the deeper reason it's unique: 2,420 breaks into exactly one set of primes.

2,420 = 22 × 5 × 112

Every 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 = 2520. The prime factorization tells you the only breakdown - there's no second option hiding.

Answer: C

Mehta17
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
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