Difference between revisions of "1951 AHSME Problems/Problem 19"
(→Solution 2) |
|||
| Line 8: | Line 8: | ||
== Solution 2 == | == Solution 2 == | ||
| − | <math>\overline{abcabc}\$ = </math>1001<math>\overline{abc}\ .So the answer is </math>\textbf{(E)}\$ | + | <math>\overline{abcabc}\$ = </math>1001<math>\overline{abc}\ . |
| + | So the answer is </math>\textbf{(E)}\$ | ||
~GEOMETRY-WIZARD. | ~GEOMETRY-WIZARD. | ||
Revision as of 06:58, 31 December 2023
Contents
Problem
A six place number is formed by repeating a three place number; for example, 
or 
, etc. Any number of this form is always exactly divisible by:
Solution
We can express any of these types of numbers in the form 
, where 
is a 3-digit number. Therefore, the answer is 
.
Solution 2
1001
\textbf{(E)}$
~GEOMETRY-WIZARD.
See Also
| 1951 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 18 |
Followed by Problem 20 | |
| 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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
