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08 Jun 2021, 02:15
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Difficulty:
15%
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Question Stats:
81% (01:09) correct
19%
(01:54)
wrong
based on 43
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If the 6-digit number 254,6bc is completely divisible by 3, find the possible values of b + c.
(A) 2
(B) 3
(C) 5
(D) 13
(E) 17
(A) 2
(B) 3
(C) 5
(D) 13
(E) 17
08 Jun 2021, 02:28
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A number is divisible by 3 if sum of all its digits is divisible by 3.
Therefore , 2+5+4+6+b+c = 17+b+c should be divisible by 3.
By putting in values from options for b+c , only 13 will suffice.
17+13=30=3+0=3 is divisible by 3.
Answer = D
Therefore , 2+5+4+6+b+c = 17+b+c should be divisible by 3.
By putting in values from options for b+c , only 13 will suffice.
17+13=30=3+0=3 is divisible by 3.
Answer = D
08 Jun 2021, 02:41
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A number is divisible by 3 if sum of all its digits is divisible by 3.
Therefore , 2+5+4+6+b+c = 17+b+c should be divisible by 3.
By putting in values from options for b+c , only 13 will suffice.
17+13=30=3+0=3 is divisible by 3.
Answer = D
Therefore , 2+5+4+6+b+c = 17+b+c should be divisible by 3.
By putting in values from options for b+c , only 13 will suffice.
17+13=30=3+0=3 is divisible by 3.
Answer = D
