When
the eggs are taken out of a basket 2, 3, 4, 5, 6, 7 at a time, the
remainders 1, 2, 3, 4, 5 and 0. How many eggs were in the basket?

Apparently such problems amused both the high and the low at that time. The divisors are not relatively prime, and relative primality is not a condition for solvability.

When the eggs were taken out 2 at a time, the remainder was 1. We can write an equation:

2k+1 [1.1]

to represent the solutions.

When this equation is divided by 3, the remainder is 2, so

(2k+1)/3 gives a remainder 2

By probing we find this value is k=2, and

2k+1=5

5 is a solution so far to division by 2 and 3. If we write a new equation:

5+2·3·k=5+6·k, [1.2]

This equation results in 1 when divided by 2 and 2 when divided by 3, for all integers k. We now require that:

(5+6·k)/4 [1.3]

gives a remainder 3, simplifying 1.3:

1+k+(1+2k)/4 [1.4]

Clearly, k=1, so (from 1.2)

5+6·k=11

So far we have a number 11, which divided by 2, gives 1, by 3, gives 2 and by 4, gives 3 as required.

We can preserve these results by writing the new equation:

11+3·4·k=11+12·k, [1.5]

because 3·4=12 is divisible by 2, 3, and 4, so our number will give the required results for division by 2, 3 and 4 for all integers k.

We now require that division by 5 gives a remainder 4, or

(11+12·k)/5 gives a remainder 4

(11+12·k)/5=

2+2·k+(1+2·k)/5 gives a remainder 4

So, by testing k=4, and (from 1.5)

11+12·4=59

Writing our next equation:

59+3·4·5·k=59+60·k [1.6]

because 3·4·5=60 is divisible by 2, 3, 4, and 5

We require that this gives a remainder 5 when divided by 6:

(59+60·k )/6 gives a remainder 5, which it actually does, for all integers k

Our final equation, which comes out even when divided by 7 is:

59+3·4·5·k [1.7]

because 3·4·5=60 is divisible by 2, 3, 4, 5 and 6

We require that:

(59+60·k)/7 comes out even.

8+8·k+(3+4·k)/7 [1.8]

When k=1, then 1.8 comes out even.

Our new number is therefore:

59+60=119

Therefore the minimum number of eggs in the basket is 119.

■

Let a

a

This will always give a remainder r

Suppose that a

a

Will always give a remainder r

In general, if a

a

always gives the correct remainder when divided by the respective m's, as can be verified by direct division.

In effect, 2.3 is the solution to the set of

Number Theory Contents

Ken Ward's Mathematics Pages

No script follows:

Ken's book is packed with examples and explanations that enable you to discover more than 150 techniques to speed up your arithmetic and increase your understanding of numbers. Paperback and Kindle: