Difference between revisions of "1980 AHSME Problems/Problem 1"
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| − | ==Problem== | + | ==Problem 1== |
| − | The largest whole number such that seven times the number is less than 100 is | + | The largest whole number such that seven times the number is less than <math>100</math> is |
<math>\text{(A)} \ 12 \qquad \text{(B)} \ 13 \qquad \text{(C)} \ 14 \qquad \text{(D)} \ 15 \qquad \text{(E)} \ 16</math> | <math>\text{(A)} \ 12 \qquad \text{(B)} \ 13 \qquad \text{(C)} \ 14 \qquad \text{(D)} \ 15 \qquad \text{(E)} \ 16</math> | ||
| Line 7: | Line 7: | ||
== Solution == | == Solution == | ||
| − | + | Let <math>x</math> be a whole number such that <math>7x < 100</math>. Dividing by <math>7</math> yields <math>x < \frac{100}{7} = 14\dfrac{2}{7}</math>, so <math>x = \boxed{(\textbf{C})\ 14}</math> is the maximum possible value. | |
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== See also == | == See also == | ||
{{AHSME box|year=1980|before=First question|num-a=2}} | {{AHSME box|year=1980|before=First question|num-a=2}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 00:57, 14 July 2025
Problem 1
The largest whole number such that seven times the number is less than 
is
Solution
Let 
be a whole number such that 
. Dividing by 
yields 
, so 
is the maximum possible value.
See also
| 1980 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by First question |
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 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
