If the swarm is divided in to 4 equal groups 3 bees will be left out and it translates to x = 4b + 3 for some natural number b

If the swarm is divided into 3 equal groups 2 bees will be left out and it translates to x = 3c + 2 for some natural number c.

If the swarm is divided into 2 equal groups 1 bee will be left out and it translates to x = 2d + 1 for some natural number d.

That is, in each case we have a positive integer y (in this example y takes values 5, 4, 3 and 2 respectively) which divides x and leaves remainder 'r' (in our case r is 0, 3, 2 and 1 respectively), that is smaller than y. In the process of writing above equations, unknowingly, we have used Division Algorithm.

Getting back to our puzzle. To solve it we must look for the multiples of 5, which satisfy all the conditions, because x = 5a + 0

If a number leaves remainder 1 when it is divided by 2 we must consider odd multiples only. In this case we have 5, 15, 25, 35 and 45. Similarly if we check for the remaining two conditions you will get 35.

Therefore, the swarm of bees contains 35 bees.

Let us verify the answer.

When 35 is divided by 2, the remainder is 1. That can be written as

35= 2 × 17 +1

When 35 is divided by 3, the remainder is 2. That can be written as

35 = 3 × 11 + 2

When 35 is divided by 4, the remainder is 3. That can be written

35 = 4 × 8 + 3

and when 35 is divided by 5, the remainder is '0'. That can be written as

35 = 5 × 7 + 0

Let us generalise this. For each pair of positive integers a and b (dividend and divisor respectively), we can find the whole numbers q and r (quotient and remainder respectively) satifying the relation

a = bq + r, 0 < r < b


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