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Bunuel
If a and b are integers such that a > b > 1, which of the following cannot be a multiple of either a or b?

(A) a – 1
(B) b + 1
(C) b – 1
(D) a + b
(E) ab

Given, a and b are integers & a > b > 1 implies that a and b are positive integers.

Note:- A positive multiple is always greater than or equal to the number it's a multiple of.
So, the answer option which violates our noted reasoning would be the correct answer to the question stem. (keeping in mind that a > b > 1)

Among answer options, b-1 is smaller than both a and b.

Ans. (C)
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Hi All,

We're told that A and B are integers such that A > B > 1. We're asked which of the following CANNOT be a multiple of either A or B. This question can be solved with some basic concept knowledge of the rules behind 'multiples.'

With the exception of "0", multiples are equal to - or greater than - their base number. For example, the multiples of 2 are 0, 2, 4, 6, 8, 10, 12, etc. Here, we're told that A is GREATER than B - and both of those integers are GREATER than 1. Thus, whatever A and B are, a number that is between B and 1 can NEVER be a multiple of either of those 2 variables. Looking at the answer choices, you should be able to immediately spot a value that is LESS than B.

Final Answer:

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Bunuel
If a and b are integers such that a > b > 1, which of the following cannot be a multiple of either a or b?

(A) a – 1
(B) b + 1
(C) b – 1
(D) a + b
(E) ab
\(a > b \ge 2\,\,\,{\rm{ints}}\)

\(?\,\,:\,\,\underline {{\rm{not}}} \,\,{\rm{multiple}}\,\,{\rm{of}}\,\,a,b\,\)

\({\rm{Take}}\,\,\left( {a,b} \right) = \left( {3,2} \right)\,\,\,\,\left\{ \matrix{\\
\,a - 1\,\,{\rm{is}}\,\,{\rm{a}}\,\,{\rm{multiple}}\,\,{\rm{of}}\,\,b\,\,\,\, \Rightarrow \,\,\,\,\left( A \right)\,\,\,{\rm{out}} \hfill \cr \\
\,b + 1\,\,{\rm{is}}\,\,{\rm{a}}\,\,{\rm{multiple}}\,\,{\rm{of}}\,\,a\,\,\,\, \Rightarrow \,\,\,\,\left( B \right)\,\,\,{\rm{out}} \hfill \cr \\
\,ab\,\,{\rm{is}}\,\,{\rm{a}}\,\,{\rm{multiple}}\,\,{\rm{of}}\,\,a,b\,\,\,\, \Rightarrow \,\,\,\,\left( E \right)\,\,\,{\rm{out}} \hfill \cr} \right.\)

\({\rm{Take}}\,\,\left( {a,b} \right) = \left( {4,2} \right)\,\,\,\,\left\{ {\,a + b\,\,{\rm{is}}\,\,{\rm{a}}\,\,{\rm{multiple}}\,\,{\rm{of}}\,\,b\,\,\,\, \Rightarrow \,\,\,\,\left( D \right)\,\,\,{\rm{out}}} \right.\)

Conclusion: the correct answer is (C), by exclusion.

Important: from the fact that b-1 is a POSITIVE integer less than both a and b, we are sure b-1 is not a multiple of any one of them!
(-2 is less than both 1 and 2, and -2 is a multiple of both of them. Be careful not to make wrong conclusions!)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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Bunuel
If a and b are integers such that a > b > 1, which of the following cannot be a multiple of either a or b?

(A) a – 1
(B) b + 1
(C) b – 1
(D) a + b
(E) ab

Since b - 1 is less than both a and b, it can’t be a multiple of either a or b.

Answer: C
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a>b>1

Therefore, b is greater than 1 but less than a

b can take the following values : {2,3,4,5,6,7...........}
a can take the following values : {3,4,5,6,7,8,...........}

from the options:
A. a-1 i.e 3-1 =2 is still a multiple of b (as 2 can be a value taken by b)
B. b+1 i.e 2+1 =3 is still a multiple of a (as 3 can be a value taken by a)
C. b-1 i.e 2-1 =1 this occurs neither in the set of values for b or a therefore this is the answer
D. a+b i.e 2+3 =5 this still a multiple of a (as 5 can be a value taken by a)
E. ab i.e 2*3=6 this is still a multiple of a (as 6 can be a value taken by a)
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