ASAL SAYILAR<\/p>\n
Asal say\u0131lar, 1 ve kendisinden ba\u015fka pozitif tam b\u00f6leni olmayan 1′ den b\u00fcy\u00fck tamsay\u0131lard\u0131r. En k\u00fc\u00e7\u00fck asal say\u0131, 2′ dir. 2 asal say\u0131s\u0131 d\u0131\u015f\u0131nda \u00e7ift asal say\u0131 yoktur. Yani, 2 say\u0131s\u0131 d\u0131\u015f\u0131ndaki t\u00fcm asal say\u0131lar tek say\u0131d\u0131r. Asal say\u0131lar k\u00fcmesi,<\/p>\n
{ 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, … }<\/p>\n
dir.<\/p>\n
Fermat Teoremi’ ne g\u00f6re, n asal say\u0131 olmak \u00fczere, 2n – 1 \u015feklinde yaz\u0131labilen say\u0131lar asal say\u0131d\u0131r. \u00d6rne\u011fin,<\/p>\n
22 – 1, 23 – 1, 25 – 1, 27 – 1, 211 – 1, …<\/p>\n
say\u0131lar\u0131, asal say\u0131d\u0131r.<\/p>\n
<\/p>\n
<\/p>\n
Aralar\u0131nda asal say\u0131lar:<\/p>\n
<\/p>\n
1′ den ba\u015fka pozitif ortak b\u00f6leni olmayan say\u0131lara, aralar\u0131nda asal say\u0131lar ad\u0131 verilir. Birden fazla say\u0131n\u0131n aralar\u0131nda asal olmas\u0131 i\u00e7in, bu say\u0131lar\u0131n asal say\u0131 olmas\u0131 gerekmez. Asal say\u0131lar, kesinlikle aralar\u0131nda asal say\u0131lard\u0131r. Bununla birlikte, 10 ve 81 say\u0131s\u0131 birer asal say\u0131 olmamas\u0131na ra\u011fmen, aralar\u0131nda asal say\u0131lard\u0131r. Di\u011fer taraftan, 10 ile 8 say\u0131s\u0131 birer asal say\u0131 olmamas\u0131na ra\u011fmen, 2 ortak b\u00f6lenleri oldu\u011fu i\u00e7in, aralar\u0131nda asal say\u0131lar de\u011fildir. Bir say\u0131 aralar\u0131nda asal iki say\u0131ya b\u00f6l\u00fcnebiliyorsa, bu iki say\u0131n\u0131n \u00e7arp\u0131m\u0131na da b\u00f6l\u00fcn\u00fcr.<\/p>\n
<\/p>\n
\u00d6rne\u011fin,<\/p>\n
\u2022 2, 9<\/p>\n
\u2022 10, 81<\/p>\n
\u2022 5, 29<\/p>\n
\u2022 3, 8<\/p>\n
\u2022 2, 10, 35<\/p>\n
say\u0131 gruplar\u0131, ortak tam b\u00f6lenleri olmad\u0131\u011f\u0131 i\u00e7in aralar\u0131nda asal say\u0131lard\u0131r.<\/p>\n
Asal olmayan say\u0131lara da bile\u015fik say\u0131 ad\u0131 verilir. Dolay\u0131s\u0131yla, bile\u015fik say\u0131lar\u0131n 1 ve kendisinden ba\u015fka b\u00f6lenleri vard\u0131r. \u00d6rne\u011fin, 10 say\u0131s\u0131 bir bile\u015fik say\u0131d\u0131r. \u00c7\u00fcnk\u00fc, 10 say\u0131s\u0131n\u0131n 1 ve kendisinden ba\u015fka, 2 ile 5 b\u00f6leni vard\u0131r. Buradan, asal olmayan 10 say\u0131s\u0131, birer asal say\u0131 olan 2 say\u0131s\u0131 ile 5 say\u0131s\u0131n\u0131n \u00e7arp\u0131m\u0131 olarak yaz\u0131labilir. 2 ile 5 say\u0131s\u0131na, 10 say\u0131s\u0131n\u0131n asal \u00e7arpan\u0131 veya b\u00f6leni denir. Yani, bile\u015fik bir say\u0131, asal say\u0131lar\u0131n \u00e7arp\u0131m\u0131 \u015feklinde yaz\u0131labilir.<\/p>\n
<\/p>\n
\u00d6rnek:<\/p>\n
<\/p>\n
A\u015fa\u011f\u0131daki say\u0131 gruplar\u0131ndan hangisi aralar\u0131nda asald\u0131r?<\/p>\n
a) 4, 20 b) 6, 21 c) 27, 36, 39 d) 8, 24, 36 e) 3, 5, 25<\/p>\n
<\/p>\n
\u00c7\u00f6z\u00fcm:<\/p>\n
<\/p>\n
a) 4 ile 20′ nin ortak b\u00f6leni vard\u0131r ve bu da 2 ile 4′ t\u00fcr.<\/p>\n
b) 6 ile 21′ in ortak b\u00f6leni vard\u0131r ve bu da 3′ t\u00fcr.<\/p>\n
c) 27, 36 ve 39′ un ortak b\u00f6leni vard\u0131r ve ortak b\u00f6len 3′ t\u00fcr.<\/p>\n
d) 8, 24 ve 36′ n\u0131n ortak b\u00f6leni vard\u0131r ve ortak b\u00f6len 2 ve 4′ t\u00fcr.<\/p>\n
e) 3, 5 ve 25′ in ortak b\u00f6leni yoktur. \u00c7\u00fcnk\u00fc, bu \u00fc\u00e7 say\u0131y\u0131 birden b\u00f6len 1′ den ba\u015fka say\u0131 yoktur. Dolay\u0131s\u0131yla, bu say\u0131lar aralar\u0131nda asald\u0131r.<\/p>\n
<\/p>\n
\u00d6rnek :<\/p>\n
<\/p>\n
a, b ve c birbirinden farkl\u0131 rakamlar olmak \u00fczere, ab ile bc iki basamakl\u0131 aralar\u0131nda asal say\u0131lard\u0131r. Buna g\u00f6re, ab + bc toplam\u0131n\u0131n en k\u00fc\u00e7\u00fck de\u011feri ka\u00e7t\u0131r?<\/p>\n
<\/p>\n
\u00c7\u00f6z\u00fcm:<\/p>\n
<\/p>\n
Toplam\u0131n en k\u00fc\u00e7\u00fck olmas\u0131 i\u00e7in, say\u0131lar\u0131 en k\u00fc\u00e7\u00fck almal\u0131y\u0131z. Buna g\u00f6re, ab = 21 olurken. bc = 13 olmal\u0131d\u0131r. Dolay\u0131s\u0131yla,<\/p>\n
ab + bc = 21 + 13 = 34<\/p>\n
olur.<\/p>\n
<\/p>\n
<\/p>\n
<\/p>\n
SAYILARIN ASAL \u00c7ARPANLARINA AYRILMASI<\/p>\n
<\/p>\n
Her bile\u015fik say\u0131, asal say\u0131lar\u0131n veya asal say\u0131lar\u0131n kuvvetlerinin \u00e7arp\u0131m\u0131 \u015feklinde yaz\u0131labilir. Bu i\u015flemi yapmak i\u00e7in, ilgili say\u0131n\u0131n s\u0131ras\u0131yla en k\u00fc\u00e7\u00fck asal say\u0131dan ba\u015flanarak b\u00f6l\u00fcnebilmesi ara\u015ft\u0131r\u0131l\u0131r.<\/p>\n
<\/p>\n
<\/p>\n
B\u0130R SAYMA SAYISININ TAMSAYI B\u00d6LENLER\u0130<\/p>\n
Bir sayma say\u0131s\u0131n\u0131n pozitif tamsay\u0131 b\u00f6lenlerinin say\u0131s\u0131:<\/p>\n
<\/p>\n
Herhangi bir A sayma say\u0131s\u0131n\u0131n asal \u00e7arpanlar\u0131 a, b ve c olmak \u00fczere,<\/p>\n
A = am . bn . cp<\/p>\n
\u015feklinde asal \u00e7arpanlar\u0131na ayr\u0131lm\u0131\u015f ise, A sayma say\u0131s\u0131n\u0131n pozitif tamsay\u0131 b\u00f6lenlerinin say\u0131s\u0131,<\/p>\n
( m + 1 ) . ( n + 1 ) . ( p + 1 )<\/p>\n
dir. Bu say\u0131ya, 1 ile say\u0131n\u0131n kendisi dahil edilmi\u015ftir.<\/p>\n
<\/p>\n
Bir sayma say\u0131s\u0131n\u0131n t\u00fcm tamsay\u0131 b\u00f6lenlerinin say\u0131s\u0131:<\/p>\n
<\/p>\n
Herhangi bir A sayma say\u0131s\u0131n\u0131n asal \u00e7arpanlar\u0131 a, b ve c olmak \u00fczere,<\/p>\n
A = am . bn . cp<\/p>\n
\u015feklinde asal \u00e7arpanlar\u0131na ayr\u0131lm\u0131\u015f ise, A sayma say\u0131s\u0131n\u0131n t\u00fcm tamsay\u0131 b\u00f6lenlerinin say\u0131s\u0131,<\/p>\n
2 . ( m + 1 ) . ( n + 1 ) . ( p + 1 )<\/p>\n
dir. Yani, A sayma say\u0131s\u0131n\u0131n t\u00fcm tamsay\u0131 b\u00f6lenlerinin say\u0131s\u0131, pozitif b\u00f6lenlerinin say\u0131s\u0131n\u0131n 2 kat\u0131d\u0131r. Bu say\u0131ya, 1 ile say\u0131n\u0131n kendisi dahil edilmi\u015ftir.<\/p>\n
<\/p>\n
Bir sayma say\u0131s\u0131n\u0131n pozitif tamsay\u0131 b\u00f6lenlerinin toplam\u0131:<\/p>\n
<\/p>\n
Herhangi bir A sayma say\u0131s\u0131n\u0131n asal \u00e7arpanlar\u0131 a, b ve c olmak \u00fczere,<\/p>\n
A = am . bn . cp<\/p>\n
<\/p>\n
Bu toplama, 1 ile say\u0131n\u0131n kendisi dahil edilmi\u015ftir. Bir sayma say\u0131s\u0131n\u0131n t\u00fcm tamsay\u0131 b\u00f6lenlerinin toplam\u0131 ise, s\u0131f\u0131rd\u0131r.<\/p>\n
<\/p>\n
Bir sayma say\u0131s\u0131n\u0131n pozitif tamsay\u0131 b\u00f6lenlerinin \u00e7arp\u0131m\u0131:<\/p>\n
<\/p>\n
Herhangi bir A sayma say\u0131s\u0131n\u0131n asal \u00e7arpanlar\u0131 a, b ve c olmak \u00fczere,<\/p>\n
<\/p>\n
A = am . bn . cp<\/p>\n
<\/p>\n
\u015feklinde asal \u00e7arpanlar\u0131na ayr\u0131lm\u0131\u015f ise, A sayma say\u0131s\u0131n\u0131n pozitif tamsay\u0131 b\u00f6lenlerinin \u00e7arp\u0131m\u0131,\u00fcss\u00fcn A\u2019n\u0131n pozitif tamsay\u0131 b\u00f6lenlerinin yar\u0131s\u0131 kadard\u0131r.<\/p>\n
<\/p>\n
<\/p>\n
\u00d6rnek :<\/p>\n
<\/p>\n
120 say\u0131s\u0131n\u0131n<\/p>\n
a) Ka\u00e7 tane pozitif b\u00f6leni vard\u0131r?<\/p>\n
b) Ka\u00e7 tane tamsay\u0131 b\u00f6leni vard\u0131r?<\/p>\n
c) Pozitif b\u00f6lenlerinin toplam\u0131 ka\u00e7t\u0131r?<\/p>\n
d) Pozitif b\u00f6lenlerinin \u00e7arp\u0131m\u0131 ka\u00e7t\u0131r?<\/p>\n
<\/p>\n
\u00c7\u00f6z\u00fcm:<\/p>\n
a) 120 say\u0131s\u0131n\u0131n asal \u00e7arpanlar\u0131na ayr\u0131lm\u0131\u015f \u015fekli<\/p>\n
120 = 23 . 31. 51<\/p>\n
oldu\u011fundan, pozitif b\u00f6lenlerinin say\u0131s\u0131<\/p>\n
( 3 + 1) . ( 1 + 1 ) . ( 1 + 1 ) = 4 . 2 . 2 = 16<\/p>\n
d\u0131r.<\/p>\n
b) 120 say\u0131s\u0131n\u0131n t\u00fcm b\u00f6lenlerinin say\u0131s\u0131, pozitif b\u00f6lenlerinin say\u0131s\u0131n\u0131n 2 kat\u0131 oldu\u011funa g\u00f6re,<\/p>\n
2 . 16 = 32<\/p>\n
dir.<\/p>\n
c) 120 say\u0131s\u0131n\u0131n pozitif b\u00f6lenlerinin toplam\u0131 360 olur.<\/p>\n
<\/p>\n
d) 120 say\u0131s\u0131n\u0131n pozitif b\u00f6lenlerinin \u00e7arp\u0131m\u0131 8 dir.<\/p>\n
120<\/p>\n
\u00d6rnek :<\/p>\n
<\/p>\n
500 . 5y say\u0131s\u0131n\u0131n asal olmayan 40 tane tamsay\u0131 b\u00f6leni varsa, y ka\u00e7t\u0131r?<\/p>\n
<\/p>\n
\u00c7\u00f6z\u00fcm:<\/p>\n
500 . 5y = 22 . 53 . 5y<\/p>\n
= 22 . 53 + y<\/p>\n
<\/p>\n
2 tane asal b\u00f6leni oldu\u011fundan, t\u00fcm b\u00f6lenlerinin say\u0131s\u0131,<\/p>\n
<\/p>\n
40 + 2 = 42<\/p>\n
<\/p>\n
dir. Buradan, pozitif b\u00f6lenlerinin say\u0131s\u0131, t\u00fcm b\u00f6lenlerinin say\u0131s\u0131n\u0131n yar\u0131s\u0131 oldu\u011fundan,<\/p>\n
21 = ( 2 + 1 ) . ( 3 + x + 1 )<\/p>\n
21 = 3 . ( 4 + x )<\/p>\n
21 = 12 + 3x<\/p>\n
3x = 21 – 12<\/p>\n
3x = 9<\/p>\n
x = 3<\/p>\n
olur.<\/p>\n
<\/p>\n
<\/p>\n
OBEB (ORTAK B\u00d6LENLER\u0130N EN B\u00dcY\u00dc\u011e\u00dc)<\/p>\n
<\/p>\n
OBEB, iki veya daha \u00e7ok say\u0131y\u0131 ayn\u0131 anda b\u00f6lebilen en b\u00fcy\u00fck say\u0131d\u0131r. Verilen say\u0131lar\u0131n OBEB’ ini bulmak i\u00e7in, say\u0131lar asal \u00e7arpanlar\u0131na ayr\u0131l\u0131r ve ortak asal \u00e7arpanlar\u0131n en k\u00fc\u00e7\u00fck \u00fcsleri al\u0131n\u0131r.<\/p>\n
<\/p>\n
1. Aralar\u0131nda asal iki say\u0131n\u0131n OBEB’ i 1′ dir. Yani, a ile b aralar\u0131nda asal iki say\u0131 ise,<\/p>\n
(a, b)OBEB = 1 dir.<\/p>\n
2. Ayn\u0131 zamanda, ikiden \u00e7ok say\u0131daki say\u0131lardan en az iki tanesi aralar\u0131nda asal ise, bu say\u0131lar\u0131n OBEB’ i 1′ dir. Yani, a, b, c, d, e say\u0131lar\u0131ndan a ile b aralar\u0131nda asal ise,<\/p>\n
(a, b, c, d, e)OBEB = 1 dir.<\/p>\n
3. \u0130ki veya daha fazla say\u0131n\u0131n ortak tam b\u00f6lenlerinin say\u0131s\u0131, OBEB’ inin b\u00f6lenlerinin say\u0131s\u0131na e\u015fittir.<\/p>\n
4. Ard\u0131\u015f\u0131k iki sayma say\u0131s\u0131n\u0131n OBEB’ i 1′ dir. Yani, a ile b ard\u0131\u015f\u0131k iki sayma say\u0131s\u0131 olmak \u00fczere,<\/p>\n
(a , b)OKEK = 1 dir.<\/p>\n
<\/p>\n
<\/p>\n
\u00d6rnek :<\/p>\n
18, 30, 42 say\u0131lar\u0131n\u0131n OBEB’ i ka\u00e7t\u0131r?<\/p>\n
<\/p>\n
\u00c7\u00f6z\u00fcm:<\/p>\n
<\/p>\n
18 = 2.32<\/p>\n
30 = 2.3.5<\/p>\n
42 = 2.3.7<\/p>\n
Her \u00fc\u00e7 say\u0131n\u0131n ortak asal \u00e7arpanlar\u0131n\u0131n en k\u00fc\u00e7\u00fck \u00fcsl\u00fcs\u00fc al\u0131nmal\u0131d\u0131r. Dolay\u0131s\u0131yla,<\/p>\n
(18, 30, 42)OBEB = 2.3 = 6 d\u0131r.<\/p>\n
<\/p>\n
\u00d6rnek :<\/p>\n
<\/p>\n
100 ile 120 say\u0131lar\u0131n\u0131n OBEB’ i ka\u00e7t\u0131r?<\/p>\n
<\/p>\n
\u00c7\u00f6z\u00fcm:<\/p>\n
<\/p>\n
<\/p>\n
100 = 22.52<\/p>\n
120 = 23.3.5<\/p>\n
Her iki say\u0131n\u0131n ortak asal \u00e7arpanlar\u0131n\u0131n en k\u00fc\u00e7\u00fck \u00fcsl\u00fcs\u00fc al\u0131nmal\u0131d\u0131r. Dolay\u0131s\u0131yla,<\/p>\n
(100, 120)OBEB = 22.5 = 20 dir.<\/p>\n
<\/p>\n
\u00d6rnek :<\/p>\n
6, 15 ve 29 say\u0131lar\u0131n\u0131n OBEB’ i ka\u00e7t\u0131r?<\/p>\n
<\/p>\n
\u00c7\u00f6z\u00fcm:<\/p>\n
\u0130kiden \u00e7ok say\u0131daki say\u0131lar\u0131n en az iki tanesi aralar\u0131nda asal ise, bu say\u0131lar\u0131n OBEB’ i 1 oldu\u011fundan, verilen say\u0131lardan 6 ile 29 say\u0131s\u0131 veya 15 ile 29 say\u0131s\u0131 aralar\u0131nda asal oldu\u011fu i\u00e7in<\/p>\n
(6, 15, 29)OBEB = 1<\/p>\n
dir.<\/p>\n
<\/p>\n
\u00d6rnek :<\/p>\n
<\/p>\n
100 ile 120 say\u0131lar\u0131n\u0131n ortak tam b\u00f6lenlerinin say\u0131s\u0131 ka\u00e7t\u0131r?<\/p>\n
<\/p>\n
\u00c7\u00f6z\u00fcm:<\/p>\n
(100, 120)OBEB = 22.51 = 20<\/p>\n
oldu\u011fundan, pozitif b\u00f6lenlerinin say\u0131s\u0131,<\/p>\n
( 2 + 1) . ( 1 + 1 ) = 3 . 2 = 6<\/p>\n
bulunur. Buradan, t\u00fcm b\u00f6lenlerin say\u0131s\u0131, pozitif b\u00f6lenlerin say\u0131s\u0131n\u0131n iki kat\u0131na e\u015fit oldu\u011fundan,<\/p>\n
2 . 6 = 12 olur.<\/p>\n
<\/p>\n
\u00d6rnek :<\/p>\n
<\/p>\n
Boyutlar\u0131 9 cm, 12 cm, 15 cm olan dikd\u00f6rtgenler prizmas\u0131 bi\u00e7imindeki kutunun i\u00e7erisi, bo\u015f yer kalmayacak \u015fekilde en b\u00fcy\u00fck boyutlu k\u00fcplerle doldurulmak istenmektedir. Bu kutuya ka\u00e7 tane k\u00fcp yerle\u015ftirilebilir?<\/p>\n
<\/p>\n
\u00c7\u00f6z\u00fcm:<\/p>\n
<\/p>\n
Kutu en b\u00fcy\u00fck boyutlu k\u00fcplerle doldurulmak istendi\u011finden, 9 cm, 12 cm, 15 cm say\u0131lar\u0131n\u0131n OBEB’ i bulunmal\u0131d\u0131r. Bu nedenle,<\/p>\n
(9, 12, 15)OBEB = 3 t\u00fcr. B\u00f6ylece, en b\u00fcy\u00fck boyutlu k\u00fcp\u00fcn bir kenar\u0131 = 3 cm olur. Bir kenar\u0131 3 cm olacak \u015fekilde yerle\u015ftirilebilecek k\u00fcp say\u0131s\u0131,<\/p>\n
K\u00fcp say\u0131s\u0131 = Kutunun hacmi \/ K\u00fcp\u00fcn hacmi = 9.12.15\/3.3.3 = 3.4.5 = 60<\/p>\n
tane olur.<\/p>\n
<\/p>\n
\u00d6rnek :<\/p>\n
<\/p>\n
Boyutlar\u0131 24 m ve 60 m olan dikd\u00f6rtgen \u015feklindeki bir arsan\u0131n \u00e7evresine e\u015fit aral\u0131klarla en az say\u0131da ka\u00e7 a\u011fa\u00e7 dikilebilir?<\/p>\n
<\/p>\n
\u00c7\u00f6z\u00fcm:<\/p>\n
\u0130ki a\u011fac\u0131n aras\u0131ndaki uzakl\u0131k, dikd\u00f6rtgenin boyutlar\u0131n\u0131n OBEB’ i olur. Dolay\u0131s\u0131yla,<\/p>\n
(24, 60)OBEB = 12<\/p>\n
A\u011fa\u00e7 Say\u0131s\u0131 = \u00c7evre \/ 12 = 2 . (24 + 60) \/ 12 = 84 \/ 6 = 14<\/p>\n
dir.<\/p>\n
<\/p>\n
OKEK (ORTAK KATLARIN EN K\u00dc\u00c7\u00dc\u011e\u00dc)<\/p>\n
<\/p>\n
\u0130ki veya daha \u00e7ok say\u0131n\u0131n her birine b\u00f6l\u00fcnen en k\u00fc\u00e7\u00fck say\u0131d\u0131r. Verilen iki veya daha \u00e7ok say\u0131n\u0131n OKEK’ ini bulmak i\u00e7in, say\u0131lar asal \u00e7arpanlar\u0131n\u0131n kuvvetleri cinsinden yaz\u0131l\u0131r ve ortak asal \u00e7arpanlar\u0131ndan \u00fcsleri en b\u00fcy\u00fck olanlarla ortak olmayan asal \u00e7arpanlar\u0131n\u0131n t\u00fcm\u00fc al\u0131narak \u00e7arp\u0131l\u0131r.<\/p>\n
<\/p>\n
1. Aralar\u0131nda asal say\u0131lar\u0131n OKEK’ i, bu say\u0131lar\u0131n \u00e7arp\u0131mlar\u0131na e\u015fittir. Yani, a ile b say\u0131s\u0131 aralar\u0131nda asal say\u0131lar ise,<\/p>\n
(a, b)OKEK = a . b dir.<\/p>\n
2. a ve b iki do\u011fal say\u0131 olmak \u00fczere, bu iki do\u011fal say\u0131n\u0131n OBEB’ i ile OKEK’ inin \u00e7arp\u0131m\u0131, bu iki do\u011fal say\u0131n\u0131n \u00e7arp\u0131m\u0131na e\u015fittir. Yani, a ve b do\u011fal say\u0131s\u0131 i\u00e7in<\/p>\n
a . b = (a, b)OKEK . (a, b)OBEB dir.<\/p>\n
3. a, b, c, d sayma say\u0131lar\u0131 olmak \u00fczere,<\/p>\n
(a\/c,b\/d)OKEK = (a, b)OKEK \/ (c, d)OBEB dir.<\/p>\n
4. a ve b iki do\u011fal say\u0131 olmak \u00fczere,<\/p>\n
(a, b)OKEK = x ve (a, b)OBEB = y<\/p>\n
ise, a ile b say\u0131lar\u0131n\u0131n toplam\u0131n\u0131n en b\u00fcy\u00fck de\u011feri<\/p>\n
x + y dir.<\/p>\n
5. Ard\u0131\u015f\u0131k iki sayma say\u0131s\u0131n\u0131n OKEK’ i bu iki say\u0131n\u0131n \u00e7arp\u0131m\u0131na e\u015fittir. Yani, a ile b ard\u0131\u015f\u0131k iki sayma say\u0131s\u0131 olmak \u00fczere,<\/p>\n
(a, b)OKEK = a . b dir.<\/p>\n
6. a ile b sayma say\u0131lar\u0131 olmak \u00fczere, a < b ise,<\/p>\n
(a, b)OBEB <= a <= b <= (a, b)OKEK dir.<\/p>\n
<\/p>\n
\u00d6rnek :<\/p>\n
<\/p>\n
18 ile 45 say\u0131lar\u0131n\u0131n OKEK’ ini bulunuz.<\/p>\n
<\/p>\n
\u00c7\u00f6z\u00fcm:<\/p>\n
<\/p>\n
18 = 2 . 32<\/p>\n
45 = 32 . 5<\/p>\n
oldu\u011fundan, (18, 45)OKEK = 32 . 2 . 5 = 90 olur.<\/p>\n
<\/p>\n
\u00d6rnek :<\/p>\n
<\/p>\n
a ve b do\u011fal say\u0131lar\u0131n\u0131n OKEK’ i 48 ve OBEB’ i 8 ve bu say\u0131lardan biri 16 ise, di\u011fer say\u0131 ka\u00e7t\u0131r?<\/p>\n
<\/p>\n
\u00c7\u00f6z\u00fcm:<\/p>\n
a = 16 olsun. (16, b)OKEK = 48 ve (16, b)OBEB = 8 oldu\u011funa g\u00f6re,<\/p>\n
a . b = (a, b)OKEK . (a, b)OBEB<\/p>\n
16 . b = 48 . 8<\/p>\n
b = 24<\/p>\n
bulunur.<\/p>\n
<\/p>\n
\u00d6rnek :<\/p>\n
<\/p>\n
Herhangi iki do\u011fal say\u0131n\u0131n OKEK’ i 120 ve OBEB’ i 8 oldu\u011funa g\u00f6re, bu say\u0131lar\u0131n toplam\u0131 en \u00e7ok ka\u00e7 olabilir?<\/p>\n
<\/p>\n
\u00c7\u00f6z\u00fcm:<\/p>\n
\u0130ki do\u011fal say\u0131n\u0131n toplam\u0131 en \u00e7ok bu iki say\u0131n\u0131n OBEB’ ile OKEK’ inin toplam\u0131 kadar olabilece\u011finden,<\/p>\n
<\/p>\n
120 + 8 = 128 dir.<\/p>\n
\u00d6rnek :<\/p>\n
<\/p>\n
Boyutlar\u0131 2 cm, 4 cm, 6 cm olan dikd\u00f6rtgenler prizmas\u0131 bi\u00e7imindeki kutunun i\u00e7erisi, bo\u015f yer kalmayacak \u015fekilde en k\u00fc\u00e7\u00fck boyutlu k\u00fcplerle doldurulmak istenmektedir. Bu kutuya ka\u00e7 tane k\u00fcp yerle\u015ftirilebilir?<\/p>\n
<\/p>\n
\u00c7\u00f6z\u00fcm:<\/p>\n
<\/p>\n
Kutu en k\u00fc\u00e7\u00fck boyutlu k\u00fcplerle doldurulmak istendi\u011finden, 2 cm, 4 cm, 6 cm say\u0131lar\u0131n\u0131n OKEK’ i bulunmal\u0131d\u0131r. Bu nedenle,<\/p>\n
(2, 4, 6)OKEK = 12 t\u00fcr. B\u00f6ylece, en k\u00fc\u00e7\u00fck boyutlu k\u00fcp\u00fcn bir kenar\u0131 = 12 cm olur. Bir kenar\u0131 12 cm olacak \u015fekilde yerle\u015ftirilebilecek k\u00fcp say\u0131s\u0131,<\/p>\n
K\u00fcp say\u0131s\u0131 = Kutunun hacmi \/ K\u00fcp\u00fcn hacmi = 12.12.12\/2.4.6 = 6.3.2 = 36<\/p>\n
tane olur.<\/p>\n
<\/p>\n
\u00d6rnek :<\/p>\n
<\/p>\n
a, b, c asal say\u0131lar olmak \u00fczere,<\/p>\n
x = a2 . b3 . c5 ve y = a5 . c2<\/p>\n
ise, (x, y)OBEB = ? ve (x, y)OKEK = ? bulunuz.<\/p>\n
<\/p>\n
\u00c7\u00f6z\u00fcm:<\/p>\n
(x, y)OBEB = a2 . c2 = (a . c)2<\/p>\n
(x, y)OKEK = a5 . b3 . c5 olur.<\/p>\n
<\/p>\n
\u00d6rnek :<\/p>\n
<\/p>\n
Ay\u015fe toplar\u0131n\u0131 2′ \u015fer 2′ \u015fer, 4′ er 4′ er, 6′ \u015far 6′ \u015far sayarsa, her defas\u0131nda 1 top art\u0131yor. Ay\u015fe’ nin en az ka\u00e7 topu vard\u0131r?<\/p>\n
<\/p>\n
\u00c7\u00f6z\u00fcm:<\/p>\n
Top say\u0131s\u0131 = (2, 4, 6)OKEK + 1 = 12 + 1 = 13 t\u00fcr.<\/p>\n
<\/p>\n
\u00d6rnek :<\/p>\n
<\/p>\n
2, 3, 4 say\u0131lar\u0131na b\u00f6l\u00fcnd\u00fc\u011f\u00fcnde 1 kalan\u0131n\u0131 veren en b\u00fcy\u00fck 2 basamakl\u0131 do\u011fal say\u0131 ka\u00e7t\u0131r?<\/p>\n
<\/p>\n
\u00c7\u00f6z\u00fcm:<\/p>\n
[(2, 3, 4)OKEK] . k + 1 <= 99<\/p>\n
24 . k + 1 <= 99<\/p>\n
k = 4 olur. Buradan, say\u0131<\/p>\n
24 . 4 + 1 = 96 + 1 = 97<\/p>\n
bulunur.<\/p>\n
<\/p>\n
\u00d6rnek :<\/p>\n
\u0130ki yang\u0131n sireni 5\/7, 7\/8 saat aral\u0131klarla alarm vermektedirler. Bu iki yang\u0131n sireni ayn\u0131 anda en son Cuma g\u00fcn\u00fc sabah 04.00′ de alarm verdiklerine g\u00f6re, hangi g\u00fcn saat ka\u00e7ta tekrar birlikte alarm verirler?<\/p>\n
<\/p>\n
<\/p>\n
\u00c7\u00f6z\u00fcm:<\/p>\n
<\/p>\n
Yang\u0131n sirenleri 5\/7, 7\/8 say\u0131lar\u0131n\u0131n OKEK’ lerinde ayn\u0131 anda alarm verirler. Dolay\u0131s\u0131yla,<\/p>\n
(5\/7, 7\/8)OKEK = (5, 7)OKEK \/ (7, 8)OBEB = 35 \/ 1 = 35 saat<\/p>\n
sonra tekrar alarm verirler. O halde, Cumartesi g\u00fcn\u00fc saat 15.00′ de tekrar alarm vereceklerdir.<\/p>\n
<\/p>\n
\u00d6rnek :<\/p>\n
Bir a do\u011fal say\u0131s\u0131 5\/3, 6 say\u0131lar\u0131na b\u00f6l\u00fcnd\u00fc\u011f\u00fcnde sonu\u00e7 tamsay\u0131 oldu\u011funa g\u00f6re, bu ko\u015fula uyan en k\u00fc\u00e7\u00fck a say\u0131s\u0131 ka\u00e7t\u0131r?<\/p>\n
<\/p>\n
\u00c7\u00f6z\u00fcm:<\/p>\n
5\/3 ile 6′ n\u0131n OKEK’ ini bulmal\u0131y\u0131z. Bu takdirde,<\/p>\n
(5\/3, 6)OKEK = (5, 6)OKEK \/ (3, 1)OBEB = 30 \/ 1 = 30 olur.<\/p>\n
<\/p>\n
\u00d6rnek :<\/p>\n
<\/p>\n
OKEK’ i 7 olan a ve b do\u011fal say\u0131lar\u0131n\u0131n toplamlar\u0131n\u0131n en k\u00fc\u00e7\u00fck ve en b\u00fcy\u00fck de\u011ferlerinin \u00e7arp\u0131m\u0131 ka\u00e7 olur?<\/p>\n
<\/p>\n
\u00c7\u00f6z\u00fcm:<\/p>\n
<\/p>\n
(a, b)OKEK = 7 ve say\u0131lar\u0131n farkl\u0131 olmad\u0131klar\u0131 s\u00f6ylenmedi\u011fine g\u00f6re,<\/p>\n
a = 7 ve b = 7<\/p>\n
al\u0131nabilir. Bu durumda, a ile b’ nin toplam\u0131n\u0131n en b\u00fcy\u00fck de\u011feri<\/p>\n
a + b = 7 + 7 = 14 … (1)<\/p>\n
olur. Di\u011fer taraftan,<\/p>\n
a = 1 ve b = 7 al\u0131n\u0131rsa, a ile b’ nin toplam\u0131n\u0131n en k\u00fc\u00e7\u00fck de\u011feri<\/p>\n
a + b = 1 +7 = 8 … (2)<\/p>\n
olur. Buradan, (1) ile (2) nin \u00e7arp\u0131m\u0131 14 . 8 = 112 bulunur<\/p>\n","protected":false},"excerpt":{"rendered":"
ASAL SAYILAR Asal say\u0131lar, 1 ve kendisinden ba\u015fka pozitif tam b\u00f6leni olmayan 1′ den b\u00fcy\u00fck tamsay\u0131lard\u0131r. En k\u00fc\u00e7\u00fck asal say\u0131, 2′ dir. 2 asal say\u0131s\u0131 d\u0131\u015f\u0131nda \u00e7ift asal say\u0131 yoktur. Yani, 2 say\u0131s\u0131 d\u0131\u015f\u0131ndaki t\u00fcm asal say\u0131lar tek say\u0131d\u0131r. Asal say\u0131lar k\u00fcmesi, { 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, … … Devam\u0131n\u0131 oku…<\/a><\/p>\n","protected":false},"author":1685,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[13],"tags":[14054,14437,14438],"class_list":["post-40361","post","type-post","status-publish","format-standard","hentry","category-mt","tag-asal-sayilar","tag-asal-sayilar-ders-notu","tag-asal-sayilar-nedir"],"_links":{"self":[{"href":"https:\/\/www.sorubak.com\/blog\/wp-json\/wp\/v2\/posts\/40361","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.sorubak.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.sorubak.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.sorubak.com\/blog\/wp-json\/wp\/v2\/users\/1685"}],"replies":[{"embeddable":true,"href":"https:\/\/www.sorubak.com\/blog\/wp-json\/wp\/v2\/comments?post=40361"}],"version-history":[{"count":0,"href":"https:\/\/www.sorubak.com\/blog\/wp-json\/wp\/v2\/posts\/40361\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.sorubak.com\/blog\/wp-json\/wp\/v2\/media?parent=40361"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.sorubak.com\/blog\/wp-json\/wp\/v2\/categories?post=40361"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.sorubak.com\/blog\/wp-json\/wp\/v2\/tags?post=40361"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}