Solution:
m and n are positive integers and m.n = 132. This implies the combinations for m, n are (1, 132), (2, 66), (3, 44), (4, 33), (6, 22), (11,12)
⇒ m, n = (132, 1), (66, 2), (44, 3)...... and so on.
∴ I alone is not sufficient.
II. Given |m–n|=1
This restricts the (m, n) combination to (12, 11), (11, 12). For both the options,
m+n=23
∴ II alone is sufficient.
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