人教A版數(shù)學(xué)必修三測(cè)試13　算法案例132

第2課時(shí)進(jìn)位制課后篇鞏固探究1.將二進(jìn)制數(shù)110101(2)轉(zhuǎn)化為十進(jìn)制數(shù)為()A.106 B.53 C.55 D.解析:110101(2)=1×25+1×24+0×23+1×22+0×21+1×20=53.答案:B2.把189化為三進(jìn)制數(shù),則末位數(shù)是()A.0 B.1 C.2 D.解析:則末位數(shù)是0.答案:A3.已知k進(jìn)制的數(shù)132與十進(jìn)制的數(shù)30相等,那么k等于 ()A.7或4 B.7C.4 D.都不對(duì)解析:由題意知132(k)=30,∴1×k2+3×k1+2×k0=30.∴k2+3k28=0.∴k=4或k=7(舍去).答案:C4.將八進(jìn)制數(shù)135(8)轉(zhuǎn)化為二進(jìn)制數(shù)是()A.1110101(2) B.1010101(2)C.111001(2) D.1011101(2)解析:135(8)=1×82+3×81+5×80=93.由除2取余法知93=1011101(2),故選D.答案:D5.已知10b1(2)=a02(3),則a+b的值為()A.0 B.1C.2 D.3解析:10b1(2)=1×23+b×2+1=2b+9,a02(3)=a×32+2=9a+2,∴2b+9=9a+2,即9a2∵a∈{1,2},b∈{0,1},∴當(dāng)a=1,b=1時(shí)符合題意,當(dāng)a=2,b=112時(shí)不合題意,∴a=1,b=1,得a+b=2,故選C答案:C6.計(jì)算機(jī)中常用的十六進(jìn)制是逢16進(jìn)1的數(shù)制,采用數(shù)字0—9和字母A—F共16個(gè)記數(shù)符號(hào),這些符號(hào)與十進(jìn)制的數(shù)的對(duì)應(yīng)關(guān)系如下表:十六進(jìn)制0123456789ABCDEF十進(jìn)制0123456789101112131415例如,用十六進(jìn)制表示E+D=1B,則A×B=()A.6E B.72C.5F D.解析:A×B=10×11=110,因?yàn)?10=6×16+14,所以將110化為十六進(jìn)制數(shù)為6E.答案:A7.(2017湖北孝感期末)將二進(jìn)制數(shù)11010(2)化為八進(jìn)制數(shù)為.

解析:二進(jìn)制數(shù)11010(2)=1×24+1×23+0×22+1×21+0×20=26.∵26÷8=3……2,3÷8=0……3,∴26(10)=32(8).答案:32(8)8.已知三個(gè)數(shù)12(16),25(7),33(4),將它們按由小到大的順序排列為.

解析:將三個(gè)數(shù)都化為十進(jìn)制數(shù).12(16)=1×16+2=18,25(7)=2×7+5=19,33(4)=3×4+3=15,所以33(4)<12(16)<25(7).答案:33(4)<12(16)<25(7)9.若六進(jìn)制數(shù)1m05(6)(m為正整數(shù))化為十進(jìn)制數(shù)為293,則m=.解析:因?yàn)?m05(6)=1×63+m×62+0×61+5×60=221+36m=293,所以m=答案:210.1101(2)+1011(2)=(用二進(jìn)制數(shù)表示).

解析:1101(2)=1×23+1×22+0×21+1×20=13;1011(2)=1×23+0×22+1×21+1×20=11,則1101(2)+1011(2)=24=11000(2).答案:11000(2)11.把三進(jìn)制數(shù)2101211(3)轉(zhuǎn)化為八進(jìn)制的數(shù).解:先將三進(jìn)制化為十進(jìn)制,再將十進(jìn)制化為八進(jìn)制.2101211(3)=2×36+1×35+0×34+1×33+2×32+1×31+1×30=1458+243+27+18+3+1=1750,所以2101211(3)=3326(8).12.導(dǎo)學(xué)號(hào)38094017已知175(r)=125(10),求在這種進(jìn)制里的數(shù)76(r)應(yīng)記成十進(jìn)