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When n is divided by 14 the remainder is 10. What is the remainder when n is divided by 7? The correct answer is 3. Shouldn't it be 20 instead? Hmmmm . . .

2006-12-07 15:34:06 · 6 answers · asked by Anonymous in Science & Mathematics Mathematics

6 answers

When a number is divided by 7, the remainder is always less than 7!

"When n is divided by 14 the remainder is 10. " n = 14x+10 for some x. (think about this if you don't know why).

divide n by 7:
n/7=(14x+10)/7=(14x+7+3)/7=(14x+7)/7+3/7. clearly, 14x+7 is divisible by 7, because it equals 7(2x+1). the remainder is 3, as you can see from the term 3/7.

You can try an example. When you divide 24 by 14, the remainder is 10, so let n be 24. Divide 24 by 7.. 24=21+3, so the answer is 3 remainder 3.

2006-12-07 15:36:24 · answer #1 · answered by need help! 3 · 0 0

firstly the remainder should always be less than the divisor.
Now for the reasoning :- n can be expressed as 14*y+10. OK,now if you understand that we divide 14*y+10 by 7. we get 2y+10/7 which is 2y+1+3/7. or in short 7 multiplied by 2y+1 and remainder 3 added is = n. QED

2006-12-07 15:44:26 · answer #2 · answered by balstoall 2 · 0 0

remainder should be less than 7 :)

10=7+3, that's where the answer comes from

2006-12-07 15:36:27 · answer #3 · answered by Anonymous · 0 0

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2016-10-17 23:27:01 · answer #4 · answered by ? 4 · 0 0

n mod 14 = 10
n = 14*k +10

n mod 7 = 3
n = 7*j + 3

14k + 10 = 7j + 3
7(2k-j) = -7
2k-j = -1
2k = j-1 , and both are integers.
try j = 5, k= 2
n=38, n=38 <---------------that's one solution

and
j=3, k= 1
n=24, n=24 <-------------- another solution.

So, there are multiple solutions, I've shown 2.

Odd multiples of 7 plus 3 will do:
n=(24,38, 52, .....) are solutions
n= (2k+1)*7 for k an integer.


If you like this kind of thing study number theory.

2006-12-07 15:43:27 · answer #5 · answered by modulo_function 7 · 0 0

yea it should be 20, 3 must be wrong.

2006-12-07 15:36:33 · answer #6 · answered by leksa27 2 · 0 1

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