What is the remainder when positive integer t is divided by

Somewhat different way.......

LCM MODEL 1 :- Any Number N which when divided by p, q, r leaving the same remainder s in each case. The Number will be of the form N = k(LCM of p, q, r) + s, where k is any non negative integer.

S1 : t = 4q + 1 --------> clearly insufficient

S2 : t = 3q + 1 --------> clearly insufficient

S1 + S2 : Applying above Model --------> t = k(LCM of 4 and 3) + 1 ---------> t = 12k + 1 ---------> t = 1, 13, 25, 37, 49 .....
Multiple Answers, Hence Insufficient

Choice E

Hope that helps!
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Number picking:
Stmt 1: 1,5,9,13,17,21 and R: 1,0,4,3,2,1 :- looks like a pattern might emerge here, but nothing else - Insufficient
Stmt 2: 1,4,10,13,16 and R: 1,4,0,3,1 :- again a pattern might emerge, but nothing else - Insufficient.
Both: 13,25 R: 3,0 :- Insufficient

Any body has a quick algebra for this?

On the second thought, I think algebra will go like this:

1: The number would be 4K+1, so (4K + 1)/5 would be 4R*K + R (R means remainder)
So if K is 1 then it will be 4R*1+R = 5R => R=0
if K is 2 then it will be 4R*2 + R = 9R => R = 4 ----- Insuficient

2: The number will be 3K + 1 so (3K + 1)/5 will be 3R*K + R
So if K is 1 then it will be 3R*1 + R = 4R => R = 4
if K is 2 then it will be 3R*2 + R = 7R => R = 2 ---------- Insufficient

Both: The number will be 12K + 1 so (12K + 1)/5 will be 2R*K + R
So if K is 1 then it will be 2R*1 + R = 3R => R = 3
if K is 2 then it will be 2R * 2 + R = 5R => R = 0 --------- Insufficient

Any body has sorter way to do it!

I believe it's like this:

t/4 => remainder 1 : n can equal: 1, 5, 9, 13, 17, 21, etc
t/3 => remainder 1 : n can equal: 1, 4, 7, 10, 13, etc.

If n = 1 => n/5 remainder 1
If n = 13 => n/5 remainder 3

So it's insufficient. E

Statement 1: t=4p+1 So the remainder could be 1,5,9,13,17,21,25.... Insufficient
Statement 2: t=3q+1 The remainder could be 1,4,7,10,13,16,19... Insufficient

1 and 2: t=12N+1 We take LCM of 3 and 4 plus firs common remainder from the lists above. Now just put in numbers instead of N.
t could be 1,13,25,37 and so on...
We and conclude from this that we don't know what is the remainder of t and the answer is E.

Although not a sound approach, here is one:

t=5A + r (What is r?)

(1).

t = 4b + 1 ..... 1 5 9 13 17 ( all give different remainders by 5) Insufficient

(2).

t = 3c + 1 ......1 4 7 10 13 16 ( all give different remainders by 5) Insufficient

Combining them series would be :

t = 12X + 13 ............ 13 , 25, .... (all give different remainders by 5) Insufficient

Hence , (E).

In S1 and S2 why you have considered Quotients as q only it can be other variables instead
and if we solve S1-S2
then , q=0, this will give t=1
so, 1/4 ............. remainder 1
1/3.............. remainder 1
and, 1/5 ........... remainder 1

On the contrary I think equations would be t= 4p+1
and, t= 3q+1

Two equations three unknowns, hence no solution.
E answer



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Narenn & @Bunnel, can you please check whether I am right ??

It is asked, t=5p+q what is the value of q?

(1) says, t=4a+1, we don’t know value of a, INSUFFICIENT.

(2) says, t=3b+1, we don’t know value of b, INSUFFICIENT.

(1) + (2) , we get 4a+1=3b+1 , here a & b has unknown value. So we can’t really decide what is the reminder.

Answer E.
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