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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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