Difference between revisions of "1967 AHSME Problems/Problem 22"
(Created page with "== Problem == For natural numbers, when <math>P</math> is divided by <math>D</math>, the quotient is <math>Q</math> and the remainder is <math>R</math>. When <math>Q</math> is d...") |
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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 == | ||
− | {{AHSME box|year=1967|num-b=21|num-a=23}} | + | {{AHSME 40p box|year=1967|num-b=21|num-a=23}} |
[[Category:Introductory Number Theory Problems]] | [[Category:Introductory Number Theory Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 01:40, 16 August 2023
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 AHSC (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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.