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14 Jun 2021, 04:15
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C
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Difficulty:
95%
(hard)
Question Stats:
42% (02:24) correct
58%
(02:16)
wrong
based on 218
sessions
History
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When 5, 8 and 12 divide a multiple of 13, they leave remainders of 3, 6 and 10, respectively. What is the least such number?
(A) 169
(B) 478
(C) 598
(D) 1,298
(E) 1,602
(A) 169
(B) 478
(C) 598
(D) 1,298
(E) 1,602
Bunuel
Go 'full-elimination' mode.
Round 1 - check with 5:
(A) 169 - when divided by 5, remainder is 4 - eliminate
(B) 478 - when divided by 5, remainder is 3 - hold
(C) 598 - when divided by 5, remainder is 3 - hold
(D) 1,298 - when divided by 5, remainder is 3 - hold
(E) 1,602 - when divided by 5, remainder is 2 - eliminate
Round 2 - check with 13 (remainder should be 0):
(B) 478 - when divided by 13, remainder is 10 - reject
(C) 598 - when divided by 13, remainder is 0 - hold
(D) 1,298 - when divided by 13, remainder is -2 (or 11) - reject
Correct option - option C.
14 Jun 2021, 08:37
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If we get a remainder of 3 when dividing by 5, our units digit must be 3 or 8. So the answer can only be B, C or D. But 478 = 390 + 88 is not a multiple of 13 (since 390 is but 88 is not), nor is 1298 = 1300 - 2 (since 1300 is but 2 is not). Only one answer remains, 598.
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