the remainder ???

[ikhlas]

Hi guise,

Please any one could teach me how can i find the formula or good way to solve reminder,, I have some questions and i like to solve these as math way ?
Please any help,
When n is divided by 7 , the remainder is 5 . What is the remainder when 2n is divided by 7 ?
the answer is :- I have to think for any number which leaves a reminder of 5 when dived by 7 but if I hadn't this idea ,, could we solve this problem in other way please ???
Another one in the attachment?
Thanks in advance for any help
Ikhlas

[scoot]

I usually do these kind of problems by taking sample numbers.

for eg.

n is div by 7 remainder is 5 , so

the equation for this is

n=7*x+5.

if u take x=1

n=12.

then 2n/7= 2*12/7= 24/7 , remainder is 3 .

HTH !!!

[ikhlas]

Originally Posted by scoot

I usually do these kind of problems by taking sample numbers.

for eg.

n is div by 7 remainder is 5 , so

the equation for this is

n=7*x+5.

if u take x=1

n=12.

then 2n/7= 2*12/7= 24/7 , remainder is 3 .

HTH !!!

thanks i get your idea and i solve the other on in the attachmnet ,, thanks for your helping .
ikhlas

[ladyp]

I realized that on some GMAT questions plugging in numbers really help and this is one of those questions. No formulas just plug in sample numbers.

[hetchkay]

n = 7f + 5
2n = 2(7f +5)

2n = 7(2f) + 10
2n = 7(2f) + 7 + 3
2n = 7(2f + 1) + 3

therefore the remainder is 3



if w,x,y,z are consecutive positive integers:
(w+x)(x+y)(y+z) is always odd (odd + even = odd, and odd * odd = odd)

an odd number divided by 2 gives remainder of 1

[ikhlas]

n = 7f + 5

2n = 2(7f +5)

2n = 7(2f) + 10
2n = 7(2f) + 7 + 3
2n = 7(2f + 1) + 3

therefore the remainder is 3
2n = 2(7f +5)
you get this formula , but if they ask us that find 2n when divided on 8 not as the same number 7 . this formula , i think will not help ?what i your opinion ?!

Ikhlas

[metodiev]

Let
w = k
x = k + 1
y = k + 2
z = k + 3

then

(2k+1)(2k+3)(2k+5) = odd x odd x odd = odd ==>remainder is 1

[ikhlas]

Originally Posted by metodiev

Let
w = k
x = k + 1
y = k + 2
z = k + 3

then

(2k+1)(2k+3)(2k+5) = odd x odd x odd = odd ==>remainder is 1

thanks , I did the same solution , but now I think I stuck here ,, ( i need any formula to help me for reminder ,, may be the pick number will not help me in the real exam )


please Have a look

n>0

the reminder when n is divided by 3 is 1 , and the reminder when n+1 is divided by 2 is 1 .

what is the reminder when n-1 is divided by 6

any help please , i pick (1) but i didnt get the answer ?!
thanks
ikhlas

[metodiev]

Originally Posted by ikhlas

n>0
the reminder when n is divided by 3 is 1 , and the reminder when n+1 is divided by 2 is 1 .

what is the reminder when n-1 is divided by 6

This is a little tricky, takes more reasoning than math skills
it says the remainder when n is divided by 3 is 1
==> n = 3k + 1
and also the remainder when n + 1 is divided by 2 is 1
n + 1 = 2m + 1 ==> n = 2m ==> n is even number
but 3k + 1 then must be even ==> 3k must be odd
==> odd*odd = odd ==> k is odd
let k = 2j+1 ==> n = 3k + 1 = 3(2j + 1) + 1 = 6j + 4
==> n = 6j + 4, but we need n - 1 so we subtract 1 and...
==> n - 1 = 6j + 4 - 1 = 6j + 3 ==> the remainder when n - 1 is divided by 6 is 3

[metodiev]

I missed the first question:
so when n is divided by 7 the remainder is 5 so
n = 7k + 5
but we need 2n so we multiply by 2
2n = 14k + 10
factor out 7...
2n = 7(2k) + 10
==> the remainder when 2n is divided by 7 is 10