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A box holds fewer than 60 buttons. How many buttons are in the box?
(1) When the buttons are grouped in fives, 3 are left over.
(2) When the buttons are grouped in sevens, 1 is left over, and the number is even.
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Correct answer
C
Let n be the count, with 0 < n < 60.
Statement (1): n ≡ 3 (mod 5) gives n in {3, 8, 13, …, 58}, many values. Not sufficient.
Statement (2): n ≡ 1 (mod 7) gives {1, 8, 15, 22, 29, 36, 43, 50, 57}; intersect "even" to get {8, 22, 36, 50}. Four values. Not sufficient.
Together: from (2), n is in {8, 22, 36, 50}. Apply (1), n ≡ 3 (mod 5): 8 mod 5 = 3 (kept), 22 mod 5 = 2 (dropped), 36 mod 5 = 1 (dropped), 50 mod 5 = 0 (dropped). So only n = 8 survives. Together sufficient; neither alone. The answer is (C).
Why neither statement alone works: (1) leaves a full residue class {3, 8, 13, …, 58}, and (2), even after the even filter, still leaves four candidates {8, 22, 36, 50}. Only the conjunction, against the bounded range 0 < n < 60, collapses to a single value. The principle: a single congruence, or a congruence plus a parity filter, under-determines n, but two independent congruences over a bounded range can pin it.
Named traps. (B) over-credits statement (2): a solver who lists {8, 22, 36, 50} may mis-read a short four-value list as if it were a single determined value. (E) is the give-up reflex: assuming two "remainder" facts can never combine to a unique value, the solver dismisses the combination without intersecting the lists. In fact, filtering {8, 22, 36, 50} through n ≡ 3 (mod 5) leaves only 8, which is exactly why (C), not (E), is correct.
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