H6. Tony multiplies together at least two consecutive positive integers. He obtains the sixdigit number $N$. The left-hand digits of $N$ are '47', and the right-hand digits of $N$ are ' 74 '.
What integers does Tony multiply together?
Solution
An integer is divisible by 4 when the number formed from the rightmost two digits is a multiple of 4 , and not otherwise. But 74 is not a multiple of 4 , so $N$ is not divisible by 4 .

However, when two even numbers are multiplied together, the result is a multiple of 4 . We conclude that Tony's list of consecutive integers does not include two even numbers.

There are two possibilities: either he has multiplied three consecutive integers 'odd', 'even', 'odd'; or he has multiplied two consecutive integers. But when two consecutive integers are multiplied together the last digit is never 4 -the only options for the last digits are $0 \times 1,1 \times 2.2 \times 3,3 \times 4,4 \times 5,5 \times 6,6 \times 7,7 \times 8,8 \times 9,9 \times 0$, so the result ends in $0,2,6,2,0,0,2,6,2$ or 0 .

We are therefore trying to find an odd integer $n$ such that $n \times(n+1) \times(n+2)=' 47 \ldots 74$ '. We also know that $n+1$ is not a multiple of 4 .

Now $81 \times 82 \times 83$ fails because it ends in 6 . It is also too big, since it is bigger than $80 \times 80 \times 80=512000$.

Also $73 \times 74 \times 75$ fails because it ends in 0 . It is also too small, since it is smaller than $75 \times 75 \times 75=421875$.

The only remaining possibility is $77 \times 78 \times 79=474474$, which works.