### Consecutive Numbers

An investigation involving adding and subtracting sets of consecutive numbers. Lots to find out, lots to explore.

### Calendar Capers

Choose any three by three square of dates on a calendar page...

### Days and Dates

Investigate how you can work out what day of the week your birthday will be on next year, and the year after...

# Common Remainder

##### Age 11 to 14 ShortChallenge Level

Finn, Kiana and Halen from Academia Britanica Cuscatleca in El Salvador, Inés from King's College Alicante in Spain and Alyssa and Trist from Greenacre Public School in Australia solved the problem using trial and improvement. Finn, Kiana and Halen said:
We tried out many different [divisors] to get a remainder of 11, but many did not work. Finally, we tried 19. ... 19 times 7 [is] 133. We immediately knew that it had a remainder of 11 because 133 + 11 = 144. After that we tried 220$\div$19. It got a remainder of 11 as well.

Alyssa and Trist used a more systematic approach:
In order to find what $n$ is, we first listed the factors of 144 and 220. After that, we knew that $n$ could not have been the factors of 144 and 220. So we made an estimate on what $n$ could be. Our first guess was 7 but since 144 divided by 7 was 20 remainder 4. The remainder 4 is way off. So we increased $n$ to 15. Which when we put that into the question, we got 144 divided by 15 equals 9 remainder 9. We increased it again to 19 and we got, 144 divided by 19 which left us with 7 remainder 11. We then checked it with 220 and thankfully got 11 remainder 11 so now we can safely say that $n$ is 19.

Carlos and Marco from Academia Britanica Cuscatleca in El Salvador refined this approach:
At first, we started to divide 220 and 114 looking for a number with 11 remainder but then we subtracted 11 from both numbers and looked for a number that could divide both 209 and 133.
They then divided 133 by different numbers until they found a common factor.

Adithya and Daanish from Hymers College in the UK, Dominic from St John's Northwood School in the UK, Su Mae from Garden International School in Malaysia, Pari from Leicester Grammar School in the UK, Aryman from Thailand, ZoÃ« from Stephen Perse Foundation School in the UK, Julian from British School Manila in Phillippenes, Kira and Emma from Wycombe High School in the UK, Peter from Durham Johnston School in the UK, Toby from Chairo Christian School in Australia and the students of Parkside Community College also looked for common factors of 133 and 209. This is Su Mae's work:

If you subtract 11 from 144 and 220, you get 133 and 209.
Now we know that both 133 and 209 must both be able to be divided by $n$ without any
reminders.

Factors of 133: 1,7,19,133
Facors of 209: 1,11,19,209
The highest common factor of 133 and 209 is 19.

And to double check, 144$\div$19=7 R 11
220$\div$19=11 R 11

Mohamed from King Solomon Academy in the UK, Nicholas from St Anthony School in the UK and Alejandra from Academia Britanica Cuscatleca also began by subtracting 11, but considered 144 before checking 220. This is Nicholas' working:
I subtracted 11 off 144 giving 133. I found the factors of 133. The first I found (beside 1 and 133) was 7 and 19. I could not use 7 because it was lower than 11 (dividing by 7 can never leave a remainder greater than 6) so I divided 209 by 19 (209 is 220-11). It divided perfectly so I knew the answer had to be 19.

This problem is taken from the UKMT Mathematical Challenges.
You can find more short problems, arranged by curriculum topic, in our short problems collection.