Cakes are sold individually for $25, in bundles of 3 for $67

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Assume a is the number of cakes sold individually, b is the number of bundles with three cakes, and c is the number of bundles with five cakes.
Then we have 25a + 67b + 97c = 2,000 and a + 3b + 5c = 100.

When we subtract the first equation from 25 times the second equation, we have 25(a + 3b + 5c) – (25a + 67b + 97c) = 25(100) - 2000, 25a + 75b + 125c - 25a - 67b - 97c = 2500 - 2000, 8b + 28c = 500, or 2b + 7c = 125.

Since we have 7c = 125 – 2b, c is an odd number, and we can put c = 2k + 1 for some integer k ≥ 0.
Then we have 2b = 125 – 7c, 2b = 125 – 7(2k + 1), 2b = 125 - 14k - 7, 2b = 118 – 14k or b = 59 – 7k > 0.

We also have a = 100 – 3b – 5c, a = 100 – 3(59 – 7k) - 5(2k + 1), a = 100 – 177 + 21k – 10k - 5, a = 11k – 82.

Since a > 0, b > 0 and c > 0, we have 11k – 82 > 0, 59 – 7k > 0 and 2k + 1 > 0.

When we put those inequalities together, we have (82/11) < k < (59/7).
Since 82/11 = 7.xxx and 59/7 = 8.xxxx, we have k = 8.

Therefore, we have a = 11k – 82 = 88 – 82 = 6.

Therefore, C is the answer.
Answer: C