Difference between revisions of "1967 AHSME Problems/Problem 22"
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== Solution == | == Solution == | ||
| − | <math>\fbox{A}</math> | + | We are given <math>P = QD + R</math> and <math>Q = D'Q' + R'</math>. |
| + | |||
| + | Plugging the second equation into the first yields: | ||
| + | |||
| + | <math>P = (D'Q' + R')D + R</math> | ||
| + | |||
| + | |||
| + | <math>P = (DD')Q' + (R'D + R)</math> | ||
| + | |||
| + | If we divide <math>P</math> by <math>DD'</math>, the quotient would be <math>Q'</math>, and the remainder would be <math>R'D + R</math>, which is option <math>\fbox{A}</math>. | ||
== See also == | == See also == | ||
Revision as of 20:16, 12 July 2019
Problem
For natural numbers, when 
is divided by 
, the quotient is 
and the remainder is 
. When is divided by 
, the quotient is 
and the remainder is 
. Then, when is divided by 
, the remainder is:
Solution
We are given 
and 
.
Plugging the second equation into the first yields:
If we divide by , the quotient would be , and the remainder would be 
, which is option 
.
See also
| 1967 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 21 |
Followed by Problem 23 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
