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# If s and t are positive integers, st + s + t cannot be

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Manager
Joined: 12 Jun 2016
Posts: 217
Location: India
WE: Sales (Telecommunications)
If s and t are positive integers, st + s + t cannot be  [#permalink]

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05 Jul 2016, 04:21
1
00:00

Difficulty:

35% (medium)

Question Stats:

71% (01:54) correct 29% (01:43) wrong based on 183 sessions

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If s and t are positive integers, st + s + t cannot be

A. 5
B. 6
C. 7
D. 8
E. 9

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Senior Manager
Joined: 02 Mar 2012
Posts: 317
Schools: Schulich '16
Re: If s and t are positive integers, st + s + t cannot be  [#permalink]

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05 Jul 2016, 06:52
take cases for(s,t)

(1,2)-5
(1,3)-7
(2,2)-8
(1,4)-9

only left is B-6
SC Moderator
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Location: India
Concentration: Strategy, General Management
GMAT 1: 200 Q1 V1
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Re: If s and t are positive integers, st + s + t cannot be  [#permalink]

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05 Jul 2016, 06:59
2
Let st + t + s = x

st + t + s + 1 = x + 1
t(s + 1) + s + 1 = x + 1
(s + 1)(t + 1) = x + 1

Minimum value of (s + 1) = 2
Minimum value of (t + 1) = 2
Hence x + 1 cannot be prime

Substitute x from the given options: 6 + 1 = 7 --> prime --> st + t + s cannot be 6

Manager
Joined: 12 Jun 2016
Posts: 217
Location: India
WE: Sales (Telecommunications)
Re: If s and t are positive integers, st + s + t cannot be  [#permalink]

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05 Jul 2016, 08:41
hsbinfy wrote:
take cases for(s,t)

(1,2)-5
(1,3)-7
(2,2)-8
(1,4)-9

only left is B-6

Thanks for the reply! One question - How did you Pick the number? Was it trial and error, some logic or practice?

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If s and t are positive integers, st + s + t cannot be  [#permalink]

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19 Nov 2017, 19:05
1
susheelh wrote:
If s and t are positive integers, st + s + t cannot be

A. 5
B. 6
C. 7
D. 8
E. 9

susheelh wrote:
hsbinfy wrote:
take cases for(s,t)

(1,2)-5
(1,3)-7
(2,2)-8
(1,4)-9

only left is B-6

Thanks for the reply! One question - How did you Pick the number? Was it trial and error, some logic or practice?

susheelh , I'm not hsbinfy , but I used exactly the same numbers.

Trial and error, with a bit of method.

I started with s = t = 2
That yields 8.

It's clear that s = t = 3 is too big (9 is the first term with those numbers, never mind adding).

s = t = 1? Too small. = 4

So, increase t:
s = 1, t = 2 ---> 5

Increase t again. Why? These end numbers are small, there is multiplication, and one factor of 1 keeps things small.

s = 1, t = 3 ----> 7
s = 1, t = 4 ----> 9

I have 8, 5, 7, and 9.

Answer by POE must be 6

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If s and t are positive integers, st + s + t cannot be &nbs [#permalink] 19 Nov 2017, 19:05
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