Difference between revisions of "1967 AHSME Problems/Problem 22"
Talkinaway (talk | contribs) (→Solution) |
m (→See also) |
||
| Line 21: | Line 21: | ||
== 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 00:40, 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. 
