Difference between revisions of "1970 AHSME Problems/Problem 34"
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<math>\text{(A) } 28\quad | <math>\text{(A) } 28\quad | ||
\text{(B) } 49\quad | \text{(B) } 49\quad | ||
| − | \text{(C) } 98\quad | + | \text{(C) } 98\quad\\ |
| − | \text{(D) an odd multiple of } 7 \text{ greater than } 49\quad | + | \text{(D) an odd multiple of } 7 \text{ greater than } 49\quad\\ |
| − | \text{(E) an even multiple of } 7 \text{ greater than } 98 | + | \text{(E) an even multiple of } 7 \text{ greater than } 98 </math> |
== Solution == | == Solution == | ||
| − | <math>\fbox{C}</math> | + | We know that 13903 minus 13511 is equivalent to 392. Additionally, 14589 minus 13903 is equivalent to 686. Since we are searching for the greatest integer that divides these three integers and leaves the same remainder, the answer resides in the greatest common factor of 686 and 392. Therefore, the answer is 98, or <math>\fbox{C}</math> |
| + | |||
| + | "Credit: Skupp3" | ||
== See also == | == See also == | ||
Latest revision as of 12:45, 16 July 2024
Problem
The greatest integer that will divide 
, 
and 
and leave the same remainder is
Solution
We know that 13903 minus 13511 is equivalent to 392. Additionally, 14589 minus 13903 is equivalent to 686. Since we are searching for the greatest integer that divides these three integers and leaves the same remainder, the answer resides in the greatest common factor of 686 and 392. Therefore, the answer is 98, or 
"Credit: Skupp3"
See also
| 1970 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 33 |
Followed by Problem 35 | |
| 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 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
