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N=abc, N with hundreds digit a, tens digit b and units digit c.

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N=abc, N with hundreds digit a, tens digit b and units digit c.  [#permalink]

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New post 22 Jan 2019, 03:13
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N=abc, N with hundreds digit a, tens digit b and units digit c. If a+b+c=20, what is the value of b?

(1) The product of digits a and c is 18.
(2) b/c=1 and b>a
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Re: N=abc, N with hundreds digit a, tens digit b and units digit c.  [#permalink]

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New post 22 Jan 2019, 04:51
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mangamma wrote:
N=abc, N with hundreds digit a, tens digit b and units digit c. If a+b+c=20, what is the value of b?

(1) The product of digits a and c is 18.

(2) b/c=1 and b>a

Kudos for each explanation



(1) The product of digits a and c is 18.
the one-digit factor pairs for 18 are (2,9) or (3,6),
if (3,6) then b = 20 - 9 = 11 which is not valid because it should be one digit
if (3,6) then b = 20 - 11 = 9 #sufficient

(2) b/c=1 and b>a
it means a < b = c
the combinations can be (2,9,9), (4,8,8), (6,7,7)
no definite value #insufficient

so A
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N=abc, N with hundreds digit a, tens digit b and units digit c.  [#permalink]

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New post 22 Jan 2019, 07:09
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(1) The product of digits a and c is 18.
the only possible pairs for 18 as product are 2,9 OR 3,6

If we have 3*6=18, then b would be 20 - (3+6) = 11 (impossible, B can only be a single digit)
Therefore 3,6 is not a valid pair of solution.
So from (i) we have only 1 possible set of values of b ie 20-(9*2)=2

(2) Now if b/c=1 and b>a
possible combinations can be 6,7,7; 4,8,8; and 2,9,9.
We do not get a definite value here. SO insufficient

so A
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N=abc, N with hundreds digit a, tens digit b and units digit c.   [#permalink] 22 Jan 2019, 07:09
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N=abc, N with hundreds digit a, tens digit b and units digit c.

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