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# Is the two-digit integer, with digits R and M (in that

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Intern
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Is the two-digit integer, with digits R and M (in that [#permalink]  05 Nov 2005, 15:08
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Is the two-digit integer, with digits R and M (in that order), a multiple of 7?
(1) R + M = 13
(2) R is divisible by 3
Manager
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the number is RM

(1) R+M
with R=4 and M=9
4+9=13
49 is divisible by 7
with R=8 and M=5
8+5=13
85 is not divisible by 7
==>INSUFF

(2)
R divisible by 3, so R can only be 3, 6, or 9
digits divisible by 7 in the 30s 60s and 90s are
35 or 63.... but M could also be 1 then 31 is not divisible by 7
==>INSUFF

(1)&(2)
numbers in the 30s, 60s, and 90s, divisible by 7 are: 35, 63, and 91
Sum of these digits never equal 13. The answer is NO

==> SUFF
VP
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C.

the only number for which both statements together is:
94

this is not divisible by 7
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duttsit wrote:
C. the only number for which both statements together is: 94. this is not divisible by 7

how about 67? it is also not divisible by 7.
VP
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HIMALAYA wrote:
duttsit wrote:
C. the only number for which both statements together is: 94. this is not divisible by 7

how about 67? it is also not divisible by 7.

thats true. incidently this one too is nto divisible by 7 so answer C remains valid. Good point.
Intern
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correct. The OA is C.
Senior Manager
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(1) Insufficient.

If R+M = 13, then we can have many combintions of R and M which may or may not be divisible by 7

(2) Insufficient

R divisible by 3 imples R = 3, 6 and 7. However, nothing is said about M

(1/2)

If R = 3, 6 and 7 implies M = 10, 7 and 6 respectively.

However the only vaild options for RM is 67 or 76 neither which are divisible by 7

The answer to this question is NO.

I pick C.
Senior Manager
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Opps typo in my last statement.

(1/2)
R = 3, 6, 9 implies M = 10, 7, 4 therefore RM = 67 and 94. Neither is divisible by 7.

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Re: DS multiple of 7 [#permalink]  09 Nov 2005, 22:05
marth750 wrote:
Is the two-digit integer, with digits R and M (in that order), a multiple of 7?
(1) R + M = 13
(2) R is divisible by 3

Let RM = 94 then RM is not multiple of 7. LET RM = 49 them RM is multiple of 7. So stmt 1 is not suff.

Let RM = 94 not multiple of 7. Let RM = 63 multiple of 7. So insufficient. (Infact sice R is divisible by 3, the tens digit can only be 3,6 or 9)

Combining both: RM can be 67 or 94. Both are not divisible by 7. Hence C.
Re: DS multiple of 7   [#permalink] 09 Nov 2005, 22:05
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