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Since we want the sum of the five integers to be as small as possible, we must start with the smallest of them such that units digit is 9 so that the greater digits are consumed in units digits and the smaller digits are consumed in tens digits. With such an arrangement, the five integers are 19, 28, 37, 48, and 50. The 5th integer cannot be 59 since 9 is already used in the 1st integer.BTGmoderatorDC wrote:Five integers between 10 and 99, inclusive, are to be formed by using each of the ten digits exactly once in such a way that the sum of the five integers is as small as possible. What is the greatest possible integer that could be among these five numbers?
A. 98
B. 91
C. 59
D. 50
E. 37
OA C
Source: Official Guide
SInce we want the sum of the 5 two-digit numbers to be as small as possible, we want the tens digits of each number to be as small as possible. Since the tens digit can't be 0, the five smallest non-zero digits are 1, 2, 3, 4, 5. Therefore, the largest possible integer is in the 50s. How we pair 0, 6, 7, 8 and 9 (as the units digits) with 1, 2, 3, 4, 5 (as the tens digits) doesn't matter. For example, the sum of 10, 26, 37, 48 and 59 is equal to the sum of 19, 28, 37, 46 and 50 (notice that either sum is 180). Therefore, the greatest possible integer is 59.BTGmoderatorDC wrote:Five integers between 10 and 99, inclusive, are to be formed by using each of the ten digits exactly once in such a way that the sum of the five integers is as small as possible. What is the greatest possible integer that could be among these five numbers?
A. 98
B. 91
C. 59
D. 50
E. 37
OA C
Source: Official Guide
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