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- #1

#### Petrus

##### Well-known member

- Feb 21, 2013

- 739

Simplify as far as possible without any calculator \(\displaystyle \frac{196707}{250971}\)

Hint:

Euclidean Algorithm

Regards,

- Thread starter Petrus
- Start date

- Thread starter
- #1

- Feb 21, 2013

- 739

Simplify as far as possible without any calculator \(\displaystyle \frac{196707}{250971}\)

Hint:

Euclidean Algorithm

Regards,

- Admin
- #2

- Mar 5, 2012

- 8,851

The Euclidean Algorithm gives us that the greatest common denominator is 6783.

Simplify as far as possible without any calculator \(\displaystyle \frac{196707}{250971}\)

Hint:

Euclidean Algorithm

Regards,

Therefore the fraction simplifies to \(\displaystyle \frac{29}{37}\).

- Thread starter
- #3

- Feb 21, 2013

- 739

Hello I like Serena,The Euclidean Algorithm gives us that the greatest common denominator is 6783.

Therefore the fraction simplifies to \(\displaystyle \frac{29}{37}\).

Thanks for joining the 'challange question'! You correctly calculated gcd and simplify! I totaly forgot about this one but I Will post full solution when I got time!

Regards

\(\displaystyle |\pi\rangle\)

- Feb 9, 2012

- 33

The sum of the digits of both numbers is a multiple of 3 so both numbers are divisible by 3. Divide them both by 3

[tex] \frac{65569}{83657} [/tex]

Clearly neither divisible by 5

break this number up like this 65569 = 065 569 , into groups of 3 digits each , now subtract, 569 - 65 = 504

This gives us at once a divisibility test for 7, 11, and 13

if 504 is divisible by 7, so is 65569

if 504 is divisible by 11 so is 65569

if 504 is divisible by 13 so is 65569

we see 7 divides 504 but 11 and 13 do not

Do the same for 83657 = 083 657 , subtract 657 - 83 = 574

It's easy to see 574 is divisible by 7 since previously 504 was divisible by 7 and this is 70 more than that. No need to test for 11 and 13 since those factors won't cancel if they are there (they are not)

Divide them both by 7.

[tex] \frac{9367}{11951} [/tex]

perform previous procedure again

367 - 9 = 358 not divisible by 7.

Note* Divisibility test for 3 did not have to be done TWICE because sum of digits was NOT 9 but in general you have to repeat the test until divisibility posibility is exhausted.

For divisibility by 17 , cut off the last digit of 9367 , multiply it by 5 , then subtract that from the truncated number.

936 - 35 = 901

If 901 is divisible by 17 then so is 9367 , it is.

Do the same for 11951,

1195 - 5 = 1190

If 1190 is divisible by 17 , then so is 11951 , it is.

Divide them both by 17

[tex] \frac{551}{703} [/tex]

apply divisibility test again for 17

55 - 5 = 50

not divisible by 17 so neither is 551

For divisibility by 19 , cut off the last digit of 551 , multiply it by 2, add that to the truncated number. 55 + 2 = 57

If 57 is ... well you get the idea , it is

Do the same for 703

70 + 6 = 76 is divisible by 19

Divide them both by 19

[tex] \frac{29}{37} [/tex]

It's quite obvious now you can stop.

- Thread starter
- #5

- Feb 21, 2013

- 739

Hello agentmulder,

The sum of the digits of both numbers is a multiple of 3 so both numbers are divisible by 3. Divide them both by 3

[tex] \frac{65569}{83657} [/tex]

Clearly neither divisible by 5

break this number up like this 65569 = 065 569 , into groups of 3 digits each , now subtract, 569 - 65 = 504

This gives us at once a divisibility test for 7, 11, and 13

if 504 is divisible by 7, so is 65569

if 504 is divisible by 11 so is 65569

if 504 is divisible by 13 so is 65569

we see 7 divides 504 but 11 and 13 do not

Do the same for 83657 = 083 657 , subtract 657 - 83 = 574

It's easy to see 574 is divisible by 7 since previously 504 was divisible by 7 and this is 70 more than that. No need to test for 11 and 13 since those factors won't cancel if they are there (they are not)

Divide them both by 7.

[tex] \frac{9367}{11951} [/tex]

perform previous procedure again

367 - 9 = 358 not divisible by 7.

Note* Divisibility test for 3 did not have to be done TWICE because sum of digits was NOT 9 but in general you have to repeat the test until divisibility posibility is exhausted.

For divisibility by 17 , cut off the last digit of 9367 , multiply it by 5 , then subtract that from the truncated number.

936 - 35 = 901

If 901 is divisible by 17 then so is 9367 , it is.

Do the same for 11951,

1195 - 5 = 1190

If 1190 is divisible by 17 , then so is 11951 , it is.

Divide them both by 17

[tex] \frac{551}{703} [/tex]

apply divisibility test again for 17

55 - 5 = 50

not divisible by 17 so neither is 551

For divisibility by 19 , cut off the last digit of 551 , multiply it by 2, add that to the truncated number. 55 + 2 = 57

If 57 is ... well you get the idea , it is

Do the same for 703

70 + 6 = 76 is divisible by 19

Divide them both by 19

[tex] \frac{29}{37} [/tex]

It's quite obvious now you can stop.

Well done and thanks for my 'challange question'! I did almost forgot about this method!

Regards,

\(\displaystyle |\pi\rangle\)