Difference between revisions of "1964 AHSME Problems/Problem 3"
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− | We can replace the letters with any numbers that satisfy this condition. It must be true for any numbers we use. Let <math>x</math> = <math>5</math> <math>y</math> = <math>2</math> <math>u</math> = <math>2</math> and <math>v</math> = <math>1</math>. | + | We can replace the letters with any numbers that satisfy this condition. It must be true for any numbers we use. Let <math>x</math> = <math>5</math> <math>y</math> = <math>2</math> <math>u</math> = <math>2</math> and <math>v</math> = <math>1</math>. Plus in you numbers and get <math>13</math> ÷ <math>2</math> <math>=</math> <math>6</math> remainder <math>1</math>. Since <math>v</math> = <math>1</math>, our answer is <math>E</math>. |
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Solution by superagh | Solution by superagh |
Revision as of 18:05, 9 June 2019
We can replace the letters with any numbers that satisfy this condition. It must be true for any numbers we use. Let =
=
=
and
=
. Plus in you numbers and get
÷
remainder
. Since
=
, our answer is
.
Solution by superagh