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## Abstract Algebra

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1. NUMBER SYSTEMS 1.What is the value of M and N respectively? If M39048458N is divisile y 8 and 11!  hereM and N are sin#le di#it inte#ers? a\$ %& 8 \$8& ' c\$'& 4 d\$5& 4 e\$4& ' Correct Answer  ( '& 4. )hoice *3\$+est of ,ivisiility y 8- If the last three di#its of a nuer are divisile y 8& then the nuer is divisile y 8. /ere& last three di#its 58N are divisile y 8 if N  4 ecause 584 is divisile y 8.+est of ,ivisiility y 11If the di#its at odd and even places of a #iven nuer are eual or differ y a nuer divisile y 11& then the #iven nuer is divisile y 11.+herefore& *M 2 9 2 4 2 4 2 8\$ ( *3 2 0 2 8 2 5 2 N\$  *M 2 5\$ should e divisile y 11! this condition is satisfied hen M  '..What is the sallest nuer y  hich 880 ust e divided in order to ae it into a  perfect suare? a\$ 3 \$ 4 c\$ 5 d\$ '88039'0 is not a perfect suare8804%0 is not a perfect suare88055%' hich is perfect suare of 4hence the ans er is option c\$53. 4 85  + 2 3383 + 4 n What s the !a #e o\$ n to %a&e t a 'er\$ect s(#are) 4.6ind the a7iu value of n such that 50&is perfectly divisile y 50n.a\$3 \$4 c\$% so #ton* 5050n50 (: 3 ; 3 ; 51 ; %1/ere % is the lar#est prie factor...<o in order to find the iniu value of =n=& it is enou#h to find the iniu po er of =%=... nd for a7iu value of =n=& find a7 po er of %...6or a7. value of n& find50%1 2 50%  % 2 1  8 >uotient onlyMa7. value of n hich is perfectly divisile y 50n is *8\$Min. value is 1 Ma7 value - 8 Min @alue - 1 Unt ,t  1.8%9' 7 3 2 8%9' 7 %%  ? a\$%3'900 \$'38800 c\$'585'0 d\$%1'%40 e\$NoneAns er- -'ton B 8%9' 7 3 2 8%9' 7 %% 8%9' 7 *3 2 %%\$ >Bef- Cy ,istriutive Da  *8%9' 7 300\$ '38800.Which of the follo in# nuers ill copletely divide *49 15 ( 1\$ ?a\$8 \$14 c\$4' d\$50 -'ton A *7 n  ( 1\$ ill e divisily y *7 2 1\$ only hen n is even.*49 15  ( 1\$  E*%  \$ 15  ( 1F  *% 30  ( 1\$& hich is divisile y *% 21\$& i.e.& 8.3.131% 2 1%13 is divisile y 10?It is divisile y 10.Gnits di#it of 131% is 3 and units di#it of 1%13 is %.<u of the units di#its& 3 and % is 10. /ence 131% 2 1%13 ends in 0 hich ipliesthat it is divisile y 10.4.What is the last di#it of 1111here  in the unit place so 10\$even%'so ans is %' ast two /,ts o\$ n#%0er en/n, n 2 4  or 8 +here is only one even t o(di#it nuer hich al ays ends in itself *last t o di#its\$ ( %' i.e.%' raised to any po er #ives the last t o di#its as %'. +herefore& our purpose is to #et %' as last t o di#its for even nuers. We no that 4 ends in %' and 10 ends in 4. Also& 4 raised to an even po er al ays ends ith %' and 4 raised to an odd po er al ays ends ith4. +herefore& 434 ill end in %' and 453 ill end in 4.1.'43' a\$ 3' \$ 4 c\$ %' d\$ None of the aove.'58' a\$ 84 \$ 89 c\$ 8% d\$ None of the aove3.54380 a\$ %% \$ %' c\$ 8' d\$ None of the aove Re%an/er 1.What is the reainder hen *8\$ H 1*'\$ divides 14*%\$ 2 14*13\$solution-/ere the divisor is ;8(1;' Which can e ritten as1';%(3;%13;%,ivident 14;%214;13Dets divide the individuallyBeainder otained fro first ter of divident ill eB>14;%13;%%;B>1413%;1%Dooin# at the second ter of the divident tells that it is copletely divided y 13;% With reainder Jero. /ence the reainder ill e %5040.If 2n is divided y 1 then the reainder is 8. And if (n is divided y 1 then the reainder is '. 6ind out the reainder hen n is divided y '.2n hen divide y 1 leaves re 8: 17282n(n hen divided y 1 leaves re ': 1y2'(n  <olvin# oth e #et '*7(y\$21 and n'*72y\$2%n ill #ive u *'*'*72y\$*7(y\$\$2%\$' .<o 3';*any thin# \$ ould leav 0 re&% hen divide ould leav 1 as re 3.What s the !a #e o\$ 44444445  88888885  44444442 + 4444443844444444 2 SM67CAT-N* 1.What is in the 00th position of 134 1344 13444 134444....?3n2n*n21\$3;1%21%;904ans4.a  ccc dddd eeeee .........What is the 10th letter? N*N21\$ *15;1'\$ 15;810Anso3.Assue that f*1\$0 and f*2n\$f*\$2f*n\$24*9n(1\$. 6or all natural nuers *Inte#ers:0\$ and n. What is the value of f*1%\$? a. 543' . 4831 c. 5508 d. 483f*1\$  0! f*\$  f*121\$  f*1\$2f*1\$24*9K1K1 H 1\$ 02024K8  3f*4\$  f*2\$  f*\$2f*\$24*9KK H 1\$ 32324K35  04f*8\$  f*424\$  f*4\$2f*4\$24*9K4K4 H 1\$ 0420424K143  980f*1'\$  f*828\$  f*8\$2f*8\$24*9K8K8 H 1\$  980298024K5%5  4'0f*1%\$  f*121'\$  f*1'\$2f*1\$24*9K1'K1 H1\$  4'0202 4K143  4834.Which of the follo in# nuers ust e added to 5'%8 to #ive a reainder of 35 hendivided y 4'0? a.955. 980 c. %9%d. '18Liven nos - 5'%8<utract the reainder i.e 35 fro the nos- 5'%8 ( 35  5'43When N ill e added to 5'43 then it should e divisile y 4'0.Cy looin# at the options e can easily eliinate - 1\$980& 3\$955& and 4\$'18 so the availale option is \$ %9%5.Bavi rites a nuer. /e sees that the first no of di#its e7ceeds 4ties the su of its di#its y 3. If the nuer is increased y 18& the result is the sae as the nuer fored y reversin# its di#its. 6ind the nuer. a\$ 35 \$5% c\$4 d\$491072y4*72y\$23!1072y10y27(18!after solvin# these t o en.s e ill #et 73! y5 i.e. 35#o throu#h options it is easy.353258358*4\$2335'.+he nuer '&1&1&&%&34 are placed in the o7es a&&c&d&e&f sho s elo in a certain order such that the su of the entries in each of the e7tree ro s and each of th  e7tree coluns*i.e& top&otto ro &left ost colun and ri#ht ost colun\$ are the sae nuer . hat is the value of ? a\$ 44\$55c\$''d\$%%6ro the #iven& 92a2214  32e2f25 : a2  e2f25so a2  12 and e2f  12'92c23  142d25 : c(d  % so c34 d %so su  9212214  '' lly for other ro and colous9a14cd3ef5%. What is the reainder hen '1%21%' is divided y %?a\$1 \$' c\$0d\$3<ol- '1% od %*%(1\$1% od %*(1\$1% od %(1 1%' od %*%;23\$' od % *3\$' od %%9 od %1reainder hen '1%21%' is divided y % is1(10*ans\$8.If N4& here  is a prie nuer #reater than & ho any different positive even divisors does n have includin# n?Ans- N  K1We no that total factors of a nuer hich is in the forat of aKKcB...  * 2 1\$.* 2 1\$. *B 2 1\$ .... * 2 1\$.*1 2 1\$  'Also odd factors of any nuer can e calculated easily y not tain#  and its po ers.<o odd factors of K1  the factors of 1  *1 2 1\$  Oven factors of the nuer  ' (   4
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