| Three Digit Number |
| X is a three digit number. The product of its first two digits is between 16 and 24, inclusive. The product of its last two digits is between 35 and 45, inclusive. If the product of all its digits is equal to 72, list all possible values of X. |
Problem Moderated by: Douglas |
| Problem Solution |
Let a, b, and c be the digits of X. Then we know the following:
ab >= 16
bc >= 35
abc = 72
Also, since the product of the digits is 72, we know none of the digits are either 5 or 7.
Since ab >= 16 and bc >= 35, then ab2c >=560.
But since abc = 72, we can divide these two equations to obtain:
b > 7
Thus the middle digit is either an 8 or a 9.
If the middle digit is 8, then the first digit has to be either 2 or 3 (to make the product fall between 16 and 24) and the last digit has to be 5 or 6 (to make the product fall between 35 and 45). So the possibilities with 8 as the middle digit are: 285, 286, 385, and 386. But none of these give a digit-product of 72.
If the middle digit is 9, then the first digit must be 2, and the last digit must be 4 or 5 (using reasoning similar to that above).
Thus, with 9 as the middle digit, the possibilities are: 294, and 295.
Since 2 x 9 x 4 = 72, the only answer is: 294
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