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24 Nov 2021, 10:00
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
45% (01:49) correct
55%
(02:10)
wrong
based on 139
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If A = abc, a three digit number where a is the hundreds digit, b is the tens digit and c is the units digit, is A divisible by 3? (Given a, b and c are different.)
(1) b, c and a, in that order, form an arithmetic progression.
(2) b = a^2 and c = 3.
(1) b, c and a, in that order, form an arithmetic progression.
(2) b = a^2 and c = 3.
02 Dec 2021, 10:17
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Bunuel
A number is divisible by 3 when the sum of its digits is divisible by 3
(1) b, c and a, in that order, form an arithmetic progression.
\( b\hspace{4mm}c\hspace{4mm} a \)
\(1 \hspace{4mm}2 \hspace{4mm}3\)
\(3 \hspace{4mm}2\hspace{4mm} 1\)
\(2 \hspace{4mm}3 \hspace{4mm}4\)
\(4\hspace{4mm} 3 \hspace{4mm}2\)
\(3\hspace{4mm} 4 \hspace{4mm}5\)
\(4\hspace{4mm} 5 \hspace{4mm}6\)
etc
Hence we can see that for various combinations of \(b, c\) and \(a \) the sum of its digit is divisible by \(3\) then this statement is SUFF.
(2) \(b = a^2 \) and \(c = 3.\)
so we have \(a \hspace{2mm}a^2\hspace{2mm} 3\)
a \(\neq 1\) because then \(a\) and \(a^2\) could become same. It is given that \( a,\) \(b\) and \(c\) are different.
if \(a=2\) then we have \(2\) \(4\) \(3\) hence this is divisible by \(3 \)
a \(\neq 3\) because then \(a\) and \(c\) would become same
a \(\ngeq 4\) because then \(a^2\) \( \geq \) \(16\) , which would not remain a single digit.
Hence the only combination possible for \(a\) \(b\) and \(c\) is \(2\) \(4\) \(3 \) and this is divisible by \(3\)
Hence SUFF.
Ans D
Hope it's clear
04 Jan 2022, 03:58
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(1) b, c and a, in that order, form an arithmetic progression.
That means we can assume: c=b+k; a= b+ 2k (where k>0 and is a natural number). Thus, a+b+c=3b+3k, which is divided by 3; therefore, A is divided by 3 -> Sufficient
(2) b = a^2 and c = 3
Remember that either a or b can be 3. Therefore, there is only one case which is (a,b)=(2,4). Thus, A=243, which divided by 3 -> Sufficient
Therefore, D is the correct answer.
That means we can assume: c=b+k; a= b+ 2k (where k>0 and is a natural number). Thus, a+b+c=3b+3k, which is divided by 3; therefore, A is divided by 3 -> Sufficient
(2) b = a^2 and c = 3
Remember that either a or b can be 3. Therefore, there is only one case which is (a,b)=(2,4). Thus, A=243, which divided by 3 -> Sufficient
Therefore, D is the correct answer.
