If n is an integer greater than 5, what is the remainder when n is

number 10 is included.

Answer = A

Consider n => 6,7,8,9,10
n/5 remainders => 1,2,3,4,0
n^2 /5 remainder= 1,4,4,1,0

Only 9 is number satisfying => Rem (9^2 /5) < Rem (9/5) { 1<4}
so A or D

lets check 2nd

n => 6,7,8,9,10
rem= 1,2,3,4,0
2n= 12,14,16,18,20
rem =2,4,1,3,0
Here we have 2 remainders which are satisfying 2nd condition => D eliminated

A is ans

While solving i was aware numbers eg 10,15,20 .... will have no impact on answer because their remainder when they or their square is divided by 5 will be 0 zero- also for their double same holds ( rem zero 0)
so i did not write in the solution .. but after you mentioned i wrote .

If an integer is divided by 5 then it has 5 possible remainders= 0;1;2;3;4
(1)
n=6 r=1; n^2=36 r=1 remainders are equal
n=7 r=2; n^2=49 r=4 4 is greater than 2, so not our case
n=8 r=3; n^2=64 r=4 4 is greater than 3, so not our case
n=9 r=4; n^2=81 r=1 4 is greater than 1 That's it!!!
n=10 r=0; n^2=100 r=0 remainders are equal
So the answer is 4, Sufficient
(2) after testing the same values as we did in previous statement we get that n can be either 8 or 9, so remainders can be 3 or 4
Insufficient

Answer A.

N can not be 5 => see red part because => If n is an integer greater than 5

Thanks! Next consecutive multiple added instead

i've noticed that you're all just taking ONE case for each remainder of n/5, and blindly trusting that \(n^{2}\)/5 (for statement 1) and 2n/5 (for statement 2) will always have the same result for ALL cases of that remainder.

more specifically:
for the case in which n/5 gives remainder = 1... you're just examining n = 6, and blindly trusting that the same result will obtain for n = 11, 16, 21, 26, etc.
if you understand the actual theory behind why this is so — AND/OR if you've worked out the algebra to prove it (which i'll do in the answer key i'm going to post in 24-36 hours) — THEN you can reliably generalize from these single cases.
otherwise that's a huge risk (...1 data point doesn't constitute a "pattern"!), and you should mitigate that risk by testing a couple of different n's for each possible remainder of n/5.

for calculating remainder of n/5, there can be only 5 possible values for r = 0,1,2,3,4 . Since we are given that n is not divisible by 5 , there are only cases r=1,2,3,4

Lets say n = 5 + x ,

1. Remainder [5+x] > Remainder [(5+x)^2]
Remainder [x] > Remainder [x^2]
Since rest of the terms are divisible by 5 and hence remainder = 0 for them.
Now since we are checking for remainder for 5. x can have only 5 unique values, because all other values beyond x=5 can again be reduced to 1,2,3,4 and 5.

This is the reason only checking for 6,7,8,9,10 is sufficient

2. Similarly for second statement, it boils down to
Remainder [2x] < Remainder [x]

^^ yeah, but, the fact that you can just look at ONE NUMBER for each possible remainder ISN'T trivial.

for instance, if you were asked about √n/5 — rather than \(n^{2}\)/5 or 2n/5 — then your reasoning would NOT work:
if n is 36, then the remainder of 36/5 is 1 ... and the remainder of √36/5 = 6/5 is also 1.
if n is 81, then the remainder of 81/5 is also 1 ... BUT the remainder of √81/5 = 9/5 is NOT 1 this time (it's 4).

be careful.

Okay Thanks

Yes, in the case of √n/5, i simply would have not been able to add the remainder terms of \sqrt{5+x}] algebraically

1)I started listing out numbers>5 and their squares, till the unit-digit of the number started repeating.
n>5
n = 6 7 8 9 10 11 12 13 14 15 16
n^2 = 36 49 64 81 100 121 144 169 196 225 256

As only unit's digit matters in remainder. Now it can be easily seen by comparing that which cases will have Rem(n^2/5) < Rem(n/5)
I see two such cases =>
Valid cases: n = 9 n^2 = 81 and n=14 and n^2 = 196
Rem in each case: 4
SUFFICIENT

2) n = 6 7 8 9 10 11 12 13 14 15 16
2n = 12 14 16 18 20 22 24 26 28 30 32
Valid cases: n=8, 2n = 16 and n=9, 2n = 18
Remainder different in each case.
INSUFFCIENT


A

Is this approach correct Ron? As I seem to not have missed out any cases, or can this be wrong too?

shuzy — 
yes, your reasoning is valid. the only issue is whether you've investigated enough cases for YOU to be convinced that the pattern is infallible.

(in general, GMAC won't give you problems with "deceptive patterns" — i.e., you won't see some cycle that repeats five, six, seven times AND THEN does something totally different. One of the main directives of the quant section is to reward pattern-recognition skills, so, any such "tricky" problem would run directly against one of the main priorities of this exam.)

Please find attached an algebraic solution for this problem. (The case-testing / pattern-recognition solution has been copiously detailed in existing posts on this thread.)

Hi..

(1) The remainder when \(n^{2}\) is divided by 5 is less than the remainder when n is divided by 5.
You are basically squaring the remainder here..
Remove all those number that have same units digit in their SQUARE - 0,1,5,6 as all of them will give SAME remainder when squared

Now, as shown above by RonPurewal that the number can be written as 5k+m - what can we infer here..
two set of numbers will behave same way when it comes to remainder..
0 and 0+5=5
1 and 1+5=6
remaining
2 and 2+5=7
3 and 3+5=8
4 and 4+5=9..
just check for only 1 in the set..
units digit 2 will give remainder 2 and 2^2 means 4, so remainder =4...4>2 so discard
units digit 3 will give remainder3 and 3^2 means 9, so remainder = 9-5=4...4>2 so discard
units digit 4 will give remainder 4 and 4^2 will give 16, so remainder = 16-15=1 here 4<1.. same will be the case in 9
in both cases remainder = 4
As all other cases discarded
Suff


(2) The remainder when 2n is divided by 5 is less than the remainder when n is divided by 5.
you are basically doubling the remainder here
remainder can be 1,2,3,4.. the remainder when the dividend is 2n will be 2,4,6-5,8-5 ......2,4,1,3 respectively
so in two cases the statement works 3 and 4
insuff

A

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