Difference between revisions of "2007 Cyprus MO/Lyceum/Problem 5"
I_like_pie (talk | contribs) |
m (+) |
||
| Line 1: | Line 1: | ||
==Problem== | ==Problem== | ||
| − | If the remainder of the division of <math>a</math> with <math>35</math> is <math>23</math>, then the remainder of the division of <math>a</math> with <math>7</math> is | + | If the [[modular arithmetic|remainder]] of the division of <math>a</math> with <math>35</math> is <math>23</math>, then the remainder of the division of <math>a</math> with <math>7</math> is |
<math> \mathrm{(A) \ } 1\qquad \mathrm{(B) \ } 2\qquad \mathrm{(C) \ } 3\qquad \mathrm{(D) \ } 4\qquad \mathrm{(E) \ } 5</math> | <math> \mathrm{(A) \ } 1\qquad \mathrm{(B) \ } 2\qquad \mathrm{(C) \ } 3\qquad \mathrm{(D) \ } 4\qquad \mathrm{(E) \ } 5</math> | ||
==Solution== | ==Solution== | ||
| − | {{ | + | <math>\displaystyle a \equiv 23 \pmod{35}</math> and <math>\displaystyle 7 | 35</math>, so <math>\displaystyle a \equiv 23 \equiv 2 \pmod{7} \Longrightarrow \mathrm{B}</math>. |
==See also== | ==See also== | ||
| − | + | {{CYMO box|year=2007|l=Lyceum|num-b=4|num-a=6}} | |
| − | + | [[Category:Introductory Algebra Problems]] | |
| − | |||
| − | |||
Latest revision as of 15:22, 6 May 2007
Problem
If the remainder of the division of 
with 
is 
, then the remainder of the division of with 
is
Solution
and 
, so 
.
See also
| 2007 Cyprus MO, Lyceum (Problems) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
| 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 | ||
