If t is a positive integer and r is the remainder when t^2 + 5t + 6 is divided by 7, what is the value of r?
1. when t is divided by 7, the remainder is 6
2. When t2 is divided by 7, the remainder is 1.
Again from my practice test
This topic has expert replies
I would say the answer is D, r =2
I got this by plugging in possible values of t in the equation.
Pls confirm if the answer is correct.
Also if any one can suggest a better way than plugging in values would appreficate it
thanks a lot
I got this by plugging in possible values of t in the equation.
Pls confirm if the answer is correct.
Also if any one can suggest a better way than plugging in values would appreficate it
thanks a lot
Last edited by ri2007 on Sun Sep 09, 2007 10:51 am, edited 1 time in total.
-
- Master | Next Rank: 500 Posts
- Posts: 460
- Joined: Sun Mar 25, 2007 7:42 am
- Thanked: 27 times
Q says t^2 + 5t + 6 is divided by 7 with remainder r
stmt 1: when t is divided by 7, the remainder is 6
so we can say let t-6 (t -reminder) = x (where x will be divisible by 7)
so t =x + 6
putting this in eqn we have (x + 6) ^ 2 + 5(x+6) + 6 / 7
i.e (x^2 +17x + 72) / 7
i.e x^2/7 + 17x/7 + 72/7
now here the first two fractions will yield an integer value (no remainder as x is divisile by 7)
so 72/7 will produce the remainder
so the remainder is 2 SUFF
stmt 2: t ^ 2 / 7 gives 1 as remainder
working with t^2 will produce a root in the above eqn
so we have
let t - a (remainder of t/7) = x
so t = x+ a
i.e t^2 = (x +a) ^2
i.e. x^2 + a^2 + 2ax
divide this by 7
so we have x ^2/ 7 will be an integer
2ax /7 wll be an int
so remainder is produced by a^2/7 and this should be 1
so possible values of "a" can be 1-6 if you check with these
u will get 1/7 will give remainder 1
& 6^2 / 7 will give remaider 1
so a can be 1 or 6
so we can say that we t is divided by 7 it will give remainder 1 or 6
so t^2 will give remainder as 1
you can check this with 8 & 13
now as we calculated for stmt 1 t/7 with remainder 6, same value is here so we can say that if a =6 then remainder for eqn is 2
now for a=1
we can say
t-1 =x
so t=x +1
put in eqn provided
i.e (x +1 )^2 +5(x+1) +6/7
solving we get
x^2/7 + 7x/7 +12/7
now here again first two fractions will yield a integer value only remaider will be added coz of 12/7 which will be 5
so we have two remainders here henec INSUFF
so ans must be A
stmt 1: when t is divided by 7, the remainder is 6
so we can say let t-6 (t -reminder) = x (where x will be divisible by 7)
so t =x + 6
putting this in eqn we have (x + 6) ^ 2 + 5(x+6) + 6 / 7
i.e (x^2 +17x + 72) / 7
i.e x^2/7 + 17x/7 + 72/7
now here the first two fractions will yield an integer value (no remainder as x is divisile by 7)
so 72/7 will produce the remainder
so the remainder is 2 SUFF
stmt 2: t ^ 2 / 7 gives 1 as remainder
working with t^2 will produce a root in the above eqn
so we have
let t - a (remainder of t/7) = x
so t = x+ a
i.e t^2 = (x +a) ^2
i.e. x^2 + a^2 + 2ax
divide this by 7
so we have x ^2/ 7 will be an integer
2ax /7 wll be an int
so remainder is produced by a^2/7 and this should be 1
so possible values of "a" can be 1-6 if you check with these
u will get 1/7 will give remainder 1
& 6^2 / 7 will give remaider 1
so a can be 1 or 6
so we can say that we t is divided by 7 it will give remainder 1 or 6
so t^2 will give remainder as 1
you can check this with 8 & 13
now as we calculated for stmt 1 t/7 with remainder 6, same value is here so we can say that if a =6 then remainder for eqn is 2
now for a=1
we can say
t-1 =x
so t=x +1
put in eqn provided
i.e (x +1 )^2 +5(x+1) +6/7
solving we get
x^2/7 + 7x/7 +12/7
now here again first two fractions will yield a integer value only remaider will be added coz of 12/7 which will be 5
so we have two remainders here henec INSUFF
so ans must be A
Regards
Samir
Samir
Ans should be A
I agree with Samir's explanantion of stmt 1.
For stmt 2 - I solved like this (I think this is faster)
t^2-1 = 7n => t^2 = 7n+1 => t = sqrt(7n+1)
Fitting this into equation given in Q we get
(t^2+6)/7 will have remainder 0. t^2 gives remainder 1. 6+1 = 7
now, we need to find what is the remainder of [5*sqrt(7n+1)]/7.. putting values which are easy to calulcate
Say, n = 5 gives 30/7 remainder 2
Say, n = 9 gives 40/7 remainder 5
So, stmt 2 is NOT SUFF
I agree with Samir's explanantion of stmt 1.
For stmt 2 - I solved like this (I think this is faster)
t^2-1 = 7n => t^2 = 7n+1 => t = sqrt(7n+1)
Fitting this into equation given in Q we get
(t^2+6)/7 will have remainder 0. t^2 gives remainder 1. 6+1 = 7
now, we need to find what is the remainder of [5*sqrt(7n+1)]/7.. putting values which are easy to calulcate
Say, n = 5 gives 30/7 remainder 2
Say, n = 9 gives 40/7 remainder 5
So, stmt 2 is NOT SUFF
-
- Master | Next Rank: 500 Posts
- Posts: 460
- Joined: Sun Mar 25, 2007 7:42 am
- Thanked: 27 times
Yes Kajcha,
Your approach would be little faster, I did not use it because I thought it would be better of to work without the square root sign (mentioned it in my post).
Your approach would be little faster, I did not use it because I thought it would be better of to work without the square root sign (mentioned it in my post).
Regards
Samir
Samir
After thinking over it again, I realised we really don't have to use sqrt at all... this approach will be even faster
t^2 + 6 will give remainder 0
We just need to calculate for 5*t/7 ... substitue values
t = 6 remainder is 2
t = 8 remainder is 5
PS: I chose 6 and 8 to satisfy stmt 2 .
t^2 + 6 will give remainder 0
We just need to calculate for 5*t/7 ... substitue values
t = 6 remainder is 2
t = 8 remainder is 5
PS: I chose 6 and 8 to satisfy stmt 2 .
-
- Master | Next Rank: 500 Posts
- Posts: 460
- Joined: Sun Mar 25, 2007 7:42 am
- Thanked: 27 times
Kajcha,
If anyone gets this kind of a problem on the GMAT he/she will definitely remember u while solving it. Thanks for these shortcuts.
If anyone gets this kind of a problem on the GMAT he/she will definitely remember u while solving it. Thanks for these shortcuts.
Regards
Samir
Samir