IPMAT/Number System•5mos
imsanskar

Someone plz verify this question

Ig value of p&q in this question contradict the divisibility rule of 3 and 11

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Comments (4)

N
noonionnogarlic
•5mos

P+Q comes out to both 7 and 18. when 18 is substituted, it violates divisibility of three.

p+q=7,

trying values of p and q:

p=1, q=6. the this GP leads to 216 as 4th term

p=6, q=1, this leads to 1/36 as 4th term. hence answer is D) Both A and B.

:)

ps: try to explain what you've already tried before posting the doubt, it helps others understand ur perspective and help you out better.

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I
imsanskar
OP•5mos

Bro if we put p+q = 7 then how that no. Divisible by 11 Along with 3?

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N
noonionnogarlic
•5mos

4p8q875 is the number. for it to be divisible by 11 sum of odd position terms - sum of even position terms must be a multiple of 11 or 0. Here, (4+8+8+5)-(p+q+7)= [0 or a multiple of 11].

let's take 11 for our first case.

so, 25 - 7 -(p+q) = 11

so -(p+q) = -7,

p+q=7

so we know at p+q=7 the number is divisble by 7.

Divisibility of 3 is sum of digits must be divisible by 3. here, 4+p+q+8+8+7+5

= p+q+ 32

as p+q = 7, here the sum of digits = 7+32= 39

39 is divisible by 3 (13*3).

hence both a and c are valid possibilities.

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I
imsanskar
OP•5mos

Thanks I am not aware about this fact

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