Difference between revisions of "1990 AJHSME Problems/Problem 1"
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Let the two three-digit numbers be <math>\overline{abc}</math> and <math>\overline{def}</math>. Their [[sum]] is equal to <math>100(a+d)+10(b+e)+(c+f)</math>. | Let the two three-digit numbers be <math>\overline{abc}</math> and <math>\overline{def}</math>. Their [[sum]] is equal to <math>100(a+d)+10(b+e)+(c+f)</math>. | ||
− | To [[minimum|minimize]] this, we need to minimize the contribution of the <math>100</math> factor, so we let <math>a=4</math> and <math>d=5</math>. Similarly, we let <math>b=6</math>, <math>e=7</math>, and then <math>c=8</math> and <math>f=9</math>. The sum is <cmath>100(9)+10(13)+(17)=1047 \ | + | To [[minimum|minimize]] this, we need to minimize the contribution of the <math>100</math> factor, so we let <math>a=4</math> and <math>d=5</math>. Similarly, we let <math>b=6</math>, <math>e=7</math>, and then <math>c=8</math> and <math>f=9</math>. The sum is <cmath>100(9)+10(13)+(17)=1047 \rightarrow \boxed{\text{C}}</cmath> |
==See Also== | ==See Also== |
Latest revision as of 16:12, 29 October 2016
Problem
What is the smallest sum of two -digit numbers that can be obtained by placing each of the six digits in one of the six boxes in this addition problem?
Solution
Let the two three-digit numbers be and . Their sum is equal to .
To minimize this, we need to minimize the contribution of the factor, so we let and . Similarly, we let , , and then and . The sum is
See Also
1990 AJHSME (Problems • Answer Key • Resources) | ||
Preceded by First Problem |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AJHSME/AMC 8 Problems and Solutions |
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