@[518068:raku1989] Buddy,one doubt.the corresponding AP is 30/12,13/12,-4/12,-21/12,-38/12 and so on. So the HP is 12/30,12/13,-12/4,-12/21,-12/38 and so on. -12/4
@[518068:raku1989] Is the question, units digit (3^a + 7^b) = 6 ? I thought it was units digit(3a+7b) =6.
@[397955:rohan_bhasker] I cross checked the Q once again, it states 3a+7b=6.
But loved the way how u deduced the answer!!
@[518068:raku1989] The Q was posted in the old PG format , so the symbols might have been lost during the transition .
But my methodology if the Q was 3a+7b :
Total Cases : 200C1 * 200C1 = 40000
Favourable Cases : If the units digit of the 3a & 7b are in the format :
(6,0), (5,1) ,(4,2) ,(3,3) ,(2,4) ,(1,5) ,(0,6)
So Favourable Cases : 20C1 * 20C1 *7 = 2800
Ans ==> 7/100
@[397955:rohan_bhasker]
the qn is 3^a +7^b=6.
I have done the same question somewhere else long time back with the same options...
@[533596:Kevin88] correct the question.....because even in the source from where u got the question, it would have given in the same format. 3a + 7b. But see the solution they have provided. u ll come to know, its 3^a + 7^b.
Hope you understood the solution i provided above
@[533596:Kevin88] There are mistakes in a number of questions on the prep thread.Ex:what are the last two digits of
65 X 29 X 37 X 63 X 71X 87 X 85
The options were a)54 b)01 c)02 d)03.
Clearly the product ends in 5.so none of the options are right
@[518068:raku1989] thanks buddy.what about the HP problem?Have u looked into my query?
@[397955:rohan_bhasker] @[533596:Kevin88]
Q1 i made the mistake.....u guys are right.....In a hurry i skipped to convert back AP to HP.
So the largest number should be 8th term
@[518068:raku1989] @[533596:Kevin88] again we have proved another answer provided on the prep thread to be wrong. cheers 
@[397955:rohan_bhasker] A fantastic initiative to boost the activeness of this thread!!
@[518068:raku1989] Could you check into my steps for the sol i obtained if the Q was 3a+7b=6 ?
The real issue comes when solutions for Q's posted in VA Prep are wrong!! 
Another gud Q i came across :
Q3) In how many ways can you write the number 210 as a product of three integers?
a) 48 b)41 c)55 d)52
@[533596:Kevin88] 210 can be factorized to 2*3*5*7. now, to write it as a product of three integers, we would have to pair 2 numbers and use the other two as they are. But, doing so doesn't give me these many ways of of writing 210 as product of two integers. Correct me if mt approach is wrong
@[397955:rohan_bhasker] do you mean to say even in prep thread they have posted 4th term as the answer?? lol!!!
I thought quant section was very well maintained there irrespective of the fact that most of the questions lack detailed solutions....
@[533596:Kevin88] Your method is perfect man!!
@[533596:Kevin88]
i am getting 14 as the answer
@[585700:KatMann] I followed the same method.. but somehow screwed up after that!!.
You even have to include negative integers, for eg: 210 = 1*-1*-210 & so on.
What did you get the total no of ordered pairs as ?
@[518068:raku1989] Didn't get you bro ? 14 isn't in the options...
Henceforth will number my Q's to avoid confusion..
Its official i suck at P&C; and Probability 
Another query from my end :
Q4) 3 cards are selected at random from a pack of 52
cards. What is the probability of getting more number of kings than queens?
a) 4032/6400 b) 4076/5203 c) 3492/4803 d) 334/8830
P.S : Hope i wont piss of the puys here by irritating them with these Q's ?
Check where i am going wrong in the concept.
210 = 2*3*5*7
let a*b*c =210
a= (2^x1)(3^x2)(5^x3)(7^x4)
b= (2^y1)(3^y2)(5^y3)(7^y4)
c= (2^z1)(3^z2)(5^z3)(7^z4)
so x1+y1+z1 =1
x2+y2+z2 =1
x3+y3+z3 =1
x4+y4+z4 =1
it has a total of 81 solutions
But we need to filter out the terms getting repeated.
ie( 5,6,7), (7,6,5) (6,7,5) all need to be considered as one
So out of 81 there will not be any terms with all three equal.
2 terms equal- (with 1) - 3 terms(210,1,1) which has to be considered as just one term
After considering negative terms -3 terms
So a total of 4 terms here with (210, 1,1)
so all terms different would have repeated 6 times
So the number of terms are (81-3)/6 =13
After considering negative terms 13*3 = 39
So a total of 52 terms here with all terms distinct
So the final ans has to be 56 right??
@[518068:raku1989] Your entire sol is correct apart from the fact that after zoning in on 13 distict pairs, considering negative terms (3c2 =3), you failed to consider the case of all postive terms too..
I.e Total No of ways each pair can be arranged is 13*4=52 ways
After including the the cases of (210,1,1);
Total No of ways would be 52+3=55.
@[533596:Kevin88]
ohh.....i missed that....(210, 1, 1) will be having only three possibilities..Got it.
@[518068:raku1989] Loved the way you obtained the total no of ordered pairs.. Btw isn't there a formula or a shortcut to determine the total no of ordered pairs ?
Any idea what would have been your answer if it was only 2 integers instead of 3 ?
In continuation with the question being discussed hope this helps...
@Kevin88 said: @raku1989 Loved the way you obtained the total no of ordered pairs.. Btw isn't there a formula or a shortcut to determine the total no of ordered pairs ?Any idea what would have been your answer if it was only 2 integers instead of 3 ?
2 integers is a straight question no???
total number of factors =16
so 8 ways of writing it as a product of two positive integers
and 16 ways of writing it as a product of two integers