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Bunuel
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10^a*K + P is divisible by 81. If K, P, a and b are positive integers 02 Dec 2022, 09:15
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38% (02:37) correct 62% (02:40) wrong based on 63 sessions

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\(10^a*K + P\) is divisible by 81. If K, P, a and b are positive integers, is \(10^b*K + P\) divisible by 15?

(1) P is the lowest prime which can be expressed in the form of 6n-1, where n is a positive integer.
(2) K is one less than P
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A
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10^a*K + P is divisible by 81. If K, P, a and b are positive integers 03 Dec 2022, 03:18
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statement 1

p is lowest prime when expressed in form of 6n-1 where n is positive int

value of p can be 5,11,17... for values of n=1,2,3...

lowest value of p is 5
given
10^b*k +p= 15int

therefore

10^b *k +p =15,30,45....

for b=1 k=2

not divisible by 15

however for p=4 b=1

divisible by 15

therefore not sufficient




as per statement 2

k is one less than p

no info about p
hence insufficient


now checking I and II

b=1 p=5 k=4

10^b*4 +5 =45


for b=2 p=5 k=4
10^2 *4 +5 =405


divisible by 15


option c correct
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10^a*K + P is divisible by 81. If K, P, a and b are positive integers 08 Dec 2022, 01:44
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Bunuel explanation sir ?
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10^a*K + P is divisible by 81. If K, P, a and b are positive integers 16 Dec 2022, 10:48
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Bunuel
\(10^a*K + P\) is divisible by 81. If K, P, a and b are positive integers, is \(10^b*K + P\) divisible by 15?

(1) P is the lowest prime which can be expressed in the form of 6n-1, where n is a positive integer.
(2) K is one less than P
Show more


If \(10^a*k + p\) is divisible by \( 81\) then it is also divisble by \(3\)

If \(10^a*k + p\) is divisble by 3 then \(10^b*k + p\) is also divisble by 3 ...(I)

Note : the value of b doesn't matter as it only adds zero's and does not alter the divisibilty by \(3\)

e.g. \(81, 801, 8001\) etc are all divisble by \(3\)

If \(10^b*k + p\) is divisible by \(15\) then it is divisible by \(3\) and \(5 \)

We already know \(10^b*k+ p\) is divisible by 3 from ..(I)

All we need to know is whether \(10^b*k + p\) is divisble by \(5\).

\(10^b*k + p\) will be divisble by \(5\) only when p is \(5\) or \(5x\)

eg. \(10+5 =15, 20+10=30, 40+5 =45\) etc

(1) P is the lowest prime which can be expressed in the form of 6n-1, where n is a positive integer.

This tells us p = \(5 \)

Hence \(10^b*k + p\) is a multiple of \(3\) that also ends with a \(5\)

\(10+5 = 15 \)

\(100+5=1005\)

\(40+5 = 45\)

\(400+5 =405\)

etc.

All such numbers are divisible by \(15\)

SUFF.

(2) K is one less than P

This doesn't tell us the value of \( p \)

\(10^2*4+5=405\) Yes
\(10^2*1+2= 102\) No

INSUFF

Ans A

Hope it helped.
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10^a*K + P is divisible by 81. If K, P, a and b are positive integers 09 Jan 2023, 15:45
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Bunuel
\(10^a*K + P\) is divisible by 81. If K, P, a and b are positive integers, is \(10^b*K + P\) divisible by 15?

(1) P is the lowest prime which can be expressed in the form of 6n-1, where n is a positive integer.
(2) K is one less than P


If \(10^a*k + p\) is divisible by \( 81\) then it is also divisble by \(3\)

If \(10^a*k + p\) is divisble by 3 then \(10^b*k + p\) is also divisble by 3 ...(I)

Note : the value of b doesn't matter as it only adds zero's and does not alter the divisibilty by \(3\)

e.g. \(81, 801, 8001\) etc are all divisble by \(3\)

If \(10^b*k + p\) is divisible by \(15\) then it is divisible by \(3\) and \(5 \)

We already know \(10^b*k+ p\) is divisible by 3 from ..(I)

All we need to know is whether \(10^b*k + p\) is divisble by \(5\).

\(10^b*k + p\) will be divisble by \(5\) only when p is \(5\) or \(5x\)

eg. \(10+5 =15, 20+10=30, 40+5 =45\) etc

(1) P is the lowest prime which can be expressed in the form of 6n-1, where n is a positive integer.

This tells us p = \(5 \)

Hence \(10^b*k + p\) is a multiple of \(3\) that also ends with a \(5\)

\(10+5 = 15 \)

\(100+5=1005\)

\(40+5 = 45\)

\(400+5 =405\)

etc.

All such numbers are divisible by \(15\)

SUFF.

(2) K is one less than P

This doesn't tell us the value of \( p \)

\(10^2*4+5=405\) Yes
\(10^2*1+2= 102\) No

INSUFF

Ans A

Hope it helped.
Show more

But 102 isn't satisfying the constraint in the question stem .
Bunuel Hi , please show some light on this . I can't find any number other than 405 that would fit into this constraint. If that's the case then answer would be D.
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10^a*K + P is divisible by 81. If K, P, a and b are positive integers 10 Jan 2023, 08:24
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conqueror98
stne
Bunuel
\(10^a*K + P\) is divisible by 81. If K, P, a and b are positive integers, is \(10^b*K + P\) divisible by 15?

(1) P is the lowest prime which can be expressed in the form of 6n-1, where n is a positive integer.
(2) K is one less than P


If \(10^a*k + p\) is divisible by \( 81\) then it is also divisble by \(3\)

If \(10^a*k + p\) is divisble by 3 then \(10^b*k + p\) is also divisble by 3 ...(I)

Note : the value of b doesn't matter as it only adds zero's and does not alter the divisibilty by \(3\)

e.g. \(81, 801, 8001\) etc are all divisble by \(3\)

If \(10^b*k + p\) is divisible by \(15\) then it is divisible by \(3\) and \(5 \)

We already know \(10^b*k+ p\) is divisible by 3 from ..(I)

All we need to know is whether \(10^b*k + p\) is divisble by \(5\).

\(10^b*k + p\) will be divisble by \(5\) only when p is \(5\) or \(5x\)

eg. \(10+5 =15, 20+10=30, 40+5 =45\) etc

(1) P is the lowest prime which can be expressed in the form of 6n-1, where n is a positive integer.

This tells us p = \(5 \)

Hence \(10^b*k + p\) is a multiple of \(3\) that also ends with a \(5\)

\(10+5 = 15 \)

\(100+5=1005\)

\(40+5 = 45\)

\(400+5 =405\)

etc.

All such numbers are divisible by \(15\)

SUFF.

(2) K is one less than P

This doesn't tell us the value of \( p \)

\(10^2*4+5=405\) Yes
\(10^2*1+2= 102\) No

INSUFF

Ans A

Hope it helped.

But 102 isn't satisfying the constraint in the question stem .
Bunuel Hi , please show some light on this . I can't find any number other than 405 that would fit into this constraint. If that's the case then answer would be D.
Show more

Dear conqueror98,

This was not an easy question and I perfectly understand your reason for getting confused.

Notice the two terms in the stem :

\(10^a*K + P\) ...(i)
\(10^b*K + P\) ...(ii)

Are they same? NO.

In (i) power of \(10\) is \(a\) and in (ii) power of \(10\) is \(b \)

Additionally:

First part talks about \(10^a*K + P\) being divisible by \(81 \)
Well as second part is asking whether \(10^b*K + P\) is divisible by \(15.\)

Does the stem say \(10^b*K + P\) is divisible by 81? NO.

So when you say \(102\) does not satisfy the constraint in the stem , it is not correct.

\(102\) does not need to be divisible \(81,\) but it definitely needs to be divisible by \(3.\)

We are trying to find whether \(10^b*K + P\) is divisible by \(15.\)

Now if you go through my first explanation, slowly. I am sure things should be clear.

If anything still remains unclear. Please let me know. Thanks.
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10^a*K + P is divisible by 81. If K, P, a and b are positive integers 08 Mar 2025, 03:00
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