using only the digit 1, 2 and 3 with at least one of each, what is the smallest number that can be created which is divisible by 8 and 9?
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using only the digit 1, 2 and 3 with at least one of each, what is the smallest number that can be created which is divisible by 8 and 9?
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Answer:
3312
Step-by-step explanation:
8 × 9 = 72
Note that a number is divisible by 72 if its last three digits is divisible by 8 and if it has a digit sum that is divisible by 9.
Now, we have to minimize the number of digits using only the digits 1, 2, and 3 with at least one of each. It must be an even number for it to be a possible multiple of 8 i.e., its unit digit must be 2. So the least possible sum of the rest of the digits is 9 - 2 = 7. In this case, we can't make a three-digit number with a digit sum equal to 9 using only the given digits.
For four-digit number, only 3312 satisfies the given conditions. Therefore, the answer is 3312.