Level 2 · Ages 8–11
Read it again any time, including while you work through the exercises.
You have 47 sweets to share among 5 friends. Everyone must get the same amount. You deal them out: 5 sweets each (25 gone), then 4 more each (20 gone)... In total, everyone gets 9 sweets, and 2 are left in your hand: not enough for one more round. You have just done a Euclidean division!
In this division, 47 is the dividend (what you share), 5 is the divisor (the number of shares), 9 is the quotient (what each person gets) and 2 is the remainder. It all fits in one equality: 47 = 5 × 9 + 2. Golden rule: the remainder must always be smaller than the divisor. If it were bigger, you could deal one more round!
To divide big numbers, set out the division: look at the left digits of the dividend first, find how many times the divisor fits in, subtract, then bring down the next digit. Repeat until the end.
And how do you check your result? Walk the path backwards: quotient × divisor + remainder must give back exactly the dividend. It is your personal error detector, use it every time!
Carry out the Euclidean division of 47 by 5.
5 × 9 = 45
I look for how many times 5 fits into 47: 5 × 9 = 45 is the closest without going over (5 × 10 = 50, too big).
47 - 45 = 2
I subtract to find the remainder: 2 is left, and 2 < 5, so I stop.
47 = 5 × 9 + 2
I check: 5 × 9 + 2 = 45 + 2 = 47. The quotient is 9 and the remainder is 2.
Result: 47 = 5 × 9 + 2: quotient 9, remainder 2
Set out the division of 138 by 6.
13 ÷ 6 : 2 fois, reste 1
I take the first two digits, 13. 6 × 2 = 12, I write 2 in the quotient and 13 - 12 = 1 is left.
j'abaisse le 8 : 18
I bring down the next digit beside the remainder: that gives 18.
18 ÷ 6 = 3, reste 0
6 × 3 = 18 exactly: I write 3 in the quotient and the remainder is 0.
6 × 23 + 0 = 138
Check: 6 × 23 = 138, everything is right.
Result: 138 = 6 × 23: quotient 23, remainder 0