Group QuestionAnswer the following question based on the information g...
Any number using all the digits will be divisible by both 3 and 9. Since the numbers have to be divisible by both 2 and 5, they must end in 0.
Any number divisible by 3 and 2 will be divisible by 6. And, if a number is divisible by 8, then it will be divisible by 4.
So, if the number ends in 0, we only need to find whether it is divisible by 7 and 8. Any number whose last three digits are divisible by 8 will be divisible by 8.
Furthermore, since the numbers end in 0, for the last three digits to be divisible by 8, the two-digit number formed by the hundreds and tens digits (in that order) must be divisible by 4.
For a number to be the smallest, the digits on the left must be smallest possible. So, consider 123456***0.
The missing digits are 7, 8 and 9. But, no two-digit numbers formed by 7, 8 and 9 are divisible by 4. So, choose a slightly larger number, swapping the 7 for the 6: 123457***0.
Now, missing digits are 6, 8 and 9, which yield two two-digit numbers divisible by 4: 68 and 96.
So, we can test 1234579680 and 1234578960 for divisibility by 7.
Continuing in this manner, choosing slightly larger number each time, we get to the solution: 1234759680.
Hence, option 3.