Remainder puzzles

Hello, and welcome to Myriora Maths! I'm thinking of a number. When I divide it by two, three, four, five or six, there's always one left over. What's the smallest it could be, bigger than one?

Here's the trick: take away the one that's left over. Now the number divides exactly by two, three, four, five and six! So first, find the smallest number they all go into.

Count in sixes, the biggest of them. Six, twelve, eighteen, and so on, up to sixty. Five only goes into thirty and sixty. Four doesn't go into thirty, but it does go into sixty. So sixty is the smallest number they all go into.

Now put the one back: sixty plus one is sixty-one. Check: sixty-one divided by four is fifteen, with one left over. Sixty-one divided by six is ten, with one left over. And it works for two, three and five as well!

Here's a twist. A number leaves two over when divided by three, three over when divided by four, and four over when divided by five. Each time, it's one short of a full group! So add one, and three, four and five all go into it: sixty. The number is sixty minus one: fifty-nine.

A common mistake is multiplying all the numbers together. Two times three times four times five times six is seven hundred and twenty, plus one is seven hundred and twenty-one. That does leave one over every time, but it isn't the smallest! Sixty-one is much smaller.

Your turn! Eggs are packed in threes, fours or sixes, and there are always two eggs left over. What's the fewest eggs there could be, more than two? You've got twenty seconds, or pause the video if you need more time. Go!

Time's up! Take away the two left over. Now three, four and six must all go into it. The smallest number they all go into is twelve. Put the two back: fourteen eggs!

Quick review. Take away the remainder. Find the smallest number all the divisors go into. Then put the remainder back. And if it's one short each time, take one away instead. See you next time!