Ig value of p&q in this question contradict the divisibility rule of 3 and 11
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Academic discussion and doubt solving for: HCF & LCM, Integral Solutions, Divisibility Rules, Factorisation, Unit Digit, Remainders.
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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.
Bro if we put p+q = 7 then how that no. Divisible by 11 Along with 3?
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.
Thanks I am not aware about this fact