The given question is a challenging GMAT 700+ level quant problem solving question testing concepts in number properties. Concepts covered: Factorials and Prime Factorisation of numbers.

Question 24: If y is the highest power of a number 'x' that can divide 101! without leaving a remainder, then for which among the following values of x will y be the highest?

- 111
- 462
- 74
- 33
- 210

INR

It is clear that all the answer options given are composite number. So, the divisor is not prime.

__Step 1__: Prime factorize the divisor, ‘x’ in this case.

__Step 2__: Compute the highest power of each of the prime factors that divides n!

__Step 3__: The highest power of x that divides n! is determined by the power of that prime factor which available in the least number.

**Key Inference**: The value of 'y' will be the highest for such an x whose highest prime factor is the smallest.

**Option A**: 111 = 3 × 37. The highest prime factor of 111 is 37.

**Option B**: 462 = 2 × 3 × 7 × 11. The highest prime factor of 462 is 11.

**Option C**: 74 = 2 × 37. The highest prime factor of 74 is 37.

**Option D**: 33 = 3 × 11. The highest prime factor of 33 is 11.

**Option E**: 210 = 2 × 3 × 5 × 7. The highest prime factor of 210 is 7.

210 is the number which has the smallest value of the highest prime factor among the 5 given options.

So, the value of y will be highest for 210.

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