Question

the difference between two numbers is 16.what can be said about the total numbers divisible by 7 that lie between these two numbers.

Answer 1

what do you think ?
for example, consider the numbers between 0 and 16

Answer 2

0,7,14.. total 3 no's

Answer 3

Let 'a' and 'b' be the two numbers, with 'a' being the smaller number and 'b' being the bigger number and b = a + 16
Consider three cases:
1) 'a' is divisible by 7. Example: a =14 and b = 30
2) 'b' is divisible by 7. Example: a = 5 and b = 21
3) Neither 'a' nor 'b' is divisible by 7. Example: a =13 and b = 29

Find the total number of numbers divisible by 7 in each interval that includes both end values.

Answer 4

Note: We did not consider the fourth case of BOTH 'a' and 'b' being divisible by 7 because that is not possible when the difference between the two numbers, which is 16, is not divisible by 7.

Answer 5

thnks

Answer 6

only one number or many numbers which lie in the interval

Answer 7

Well, if you consider the case of a = 15 and b = 31
we have only 21 and 28 divisible by 7.

Answer 8

in first two cases the total numbers divisible by 7 are 3..
but in the third case it varies from 2,3.

Answer 9

So minimum 2 and maximum 3.

Answer 10

yes

Answer 11

Nicely explained @aum unlike some other users who ask rhetorical questions from the get-go.

Answer 12

Thank you abb0t.

Answer 13

am i correct ?

Answer 14

see if below helps to conclude :
There is exactly one and only one number divisible ny \(n\) in \(n\) consecutive integers.

Answer 15

@BSwan wanna try the proof for above statement ? :)

Answer 16

i wanna lol
brb , im washing clothes

Answer 17

most proofs are done while thinking in shower or wash room :P
I thihnk this proof involves an application of DA algorithm - it could be more involved too... not sure

Answer 18

you need to prove both existence and uniqueness

Answer 19

could you rewrite the statment plz ?

Answer 20

@ganeshie8

Answer 21

rewrite ?

Answer 22

yeah :D

Answer 23

take any "n" consecutive integers,
prove that one number will always be divisible by "n"

Answer 24

ok , i think we proved that before wilson thm , right ?

Answer 25

for example :
take 5 consecutive numbers {7,8,9,10, 11}

5 | 10

Answer 26

i don't remember wilson's proof
need to look up again

Answer 27

so let a be the first number that devide n ,
then
a+1=1 mod n
a+2=2 mod n
a+3=3 mod n
.
.
.
.
a+(n-1)= n-1 mod n

a+n=0 mod n
and ect :)

Answer 28

@ganeshie8 any other method ?

Answer 29

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