Difference between revisions of "1994 AJHSME Problems/Problem 23"
Mrdavid445 (talk | contribs) (Created page with "==Problem== If <math>X</math>, <math>Y</math> and <math>Z</math> are different digits, then the largest possible <math>3-</math>digit sum for <math>\begin{tabular}{ccc} X & X &...") |
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<math>\text{(A)}\ XXY \qquad \text{(B)}\ XYZ \qquad \text{(C)}\ YYX \qquad \text{(D)}\ YYZ \qquad \text{(E)}\ ZZY</math> | <math>\text{(A)}\ XXY \qquad \text{(B)}\ XYZ \qquad \text{(C)}\ YYX \qquad \text{(D)}\ YYZ \qquad \text{(E)}\ ZZY</math> | ||
+ | |||
+ | ==Solution== | ||
+ | The sum can be rewritten as <math>113X + 10Y</math>. To get the largest possible sum, we maximize the hundreds digit, <math>X</math>. If <math>X=9</math>, the sum is a <math>4</math>-digit number, so we let <math>X=8</math> and <math>113(8)+10Y = 904+10Y</math>. To continue maxmimizing this sum, we can let <math>Y=9</math>, a different digit from <math>X</math>, and <math>904+10(9)=994</math>, which has the form <math>\boxed{\text{(D)}\ YYZ}</math>. | ||
+ | |||
+ | ==See Also== | ||
+ | {{AJHSME box|year=1994|num-b=22|num-a=24}} |
Revision as of 01:14, 23 December 2012
Problem
If , and are different digits, then the largest possible digit sum for
has the form
Solution
The sum can be rewritten as . To get the largest possible sum, we maximize the hundreds digit, . If , the sum is a -digit number, so we let and . To continue maxmimizing this sum, we can let , a different digit from , and , which has the form .
See Also
1994 AJHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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 |