Question

Instructions: A numerical machine accepts two values X and Y. At every step, it updates the values as X = X times Y, and Y = Y + 1. The machine stops as soon as X becomes greater than or equal to N.

If X = 2 and Y = 3, what should be the minimum value of N (from the options given) such that the final value of Y is 7?

• A. 300
• B. 360
• C. 720
• D. 860

Hint:

Y reaches 7 only if the machine stops exactly at step 4, which needs N strictly above 120 and at most 720.

Answer 1

The Correct Option is A

Instead of testing each option separately, it helps to first pin down exactly which range of N values makes Y stop at 7, and then just read off the smallest option inside that range.

1. Running the machine gives X values of 6, 24, 120 and 720 after 1, 2, 3 and 4 steps, with Y reaching 4, 5, 6 and 7 respectively.
2. Y becomes 7 only if the machine runs exactly 4 steps, which needs X to still be below N after step 3 (X = 120) but at or above N after step 4 (X = 720).
3. This means N must be strictly more than 120 and at most 720.
4. Checking the four options against $120 < N \le 720$: 300 fits, 360 fits, 720 fits (right at the boundary), and 860 does not fit since it is above 720.

Of the values that fit, 300 is the smallest.

So the correct answer is option A, 300.