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CHAPTER, , NUMBER OF ZEROES, , , , A, , gece of zeroes in an Expression, , gero:- zero will be formed by 2 and 5, , Bx. 10=2%x5, , 100 = 2? x 5?, , 1000 = 2° x 5°, , We can say that for 'n' number, , of zeroes at end of the product., , We need exactly, , combinations of 5 and 2, , Ex.1 Find the number of zeroes at, the end of the product:S*x7*x9x2x11, , n, , Sol. 5x 7x9x2x 11, In this product we see, Number of 2's = 1, Number of 5's = 1, Number of pair 2's and 5's = 1, :.Number of zero = 1, , Ex.2, , Find the number of zeroes at, the end of the product:- 12 x 27, x * 63 x 113 x 1250 x 24 x 650, 12 x 27x63 x 113x 1250 x 24, + x 650, , Break the numbers form of 2, and 5 multiple, , In this series 27, 63 & 11, are not multiple of 2 & 5. *, =, , +. The multiple of 2 & 5 are, 12, 1250, 24 & 650, 12=2x2x3=22x3, , , , a, , 1250=2x5x5x5x 1, 24=2«x2x2x3 3, , , , 4 x, 50, =. 6, , gf there are 7 two's and 6 fives,, Hence we shall be able to, form only 6 pairs of 2 and 5,, Hence there will be 6 zeroes, at the end of the Product of, , eve Ex numbers,, , the ™*-3 Find the number of zeroes at, the end of the product:1*3%5%7%9 0.0... 97 «99, , 6, , Sol. 1x 3«5x7x9 97 x 99, In this series the number of zero, and the end of the product is "0"., Because there is no even number, Present in this series while it is, necessary to be 2 and 5 for the, Zero, , The highest power of k that can, exactly divided n! we divide n by k,, n by k?, n by k° and so on till we get, , A, [| equal to 1 an then add Si, n n n n n, lel Le) [iL A ee], Ex.4 Find the largesr’poly, contained in, , ‘ [24]:, , Sol. 5 i, (We, cafin it further since, , 4}is hot divisible by 5%], Hence, \tHere are 28 times 5, alternate as a factor in 124!, Altetnate:, successive quotients till, , get O as the last quotient, / $1124, , 24 — >) 28 (add up all, ‘ [4 —»f the quotients), Ex.5 Find the largest power of 3, , that can divide 270!, , PS, , 270, + “ge | =90+ 30+ 10+3+1=134, , Hence, there are 134 times, 3 involved as a factor in 270!, Alternate:Divide successive quotients til] we, get O as the last quotient, , i, er of 5, , }, ef FMA + 4 = 28, a, , , , 3|270, , 3, , 3 134 (add up all the, 3 quotients), , 3, , 3, , , , Rakesh Yadav Readers Publication Pvt. Ltd., , , , %, Ex.6- Jind Jatgest power of 2, auton can contained in:, 1 x 4. 22?, , , , , , , , oO, Number of 2's = 11+5+2+ 1=19, Hence, there are 19 times 2, involved as a factor in 22!, Ex.7 Find the largest power of 5 that, , _—Tan contained in, 1, , x2x3x4, Sol., , 42, }o, , 8 —>, Hence, there are 9 times 5 involved, as a factor in 42!, , i —, Ex.8 Find the largest power of 7, at can exactly divide 777!, , 5, 5, 5, , , , , , Sol. 7|777, , 7 [TI, , 70O35— hos (add up all the, , 7 ao quotients), Thus the highest Power of 7 is 128, by which 777!, , can be divided., , Ex.9 Find the number of zeroes at, , — the end of the Product 10!, , Sol. 1OI=1*2x3x4x5x6x7x, 8x9x10, , 2x5, In this ty;, , clear that i, , , , number of 2's In this, Condition. all we need to do is, to count the number of S's, Number of 5°: 2, , Number of 2's = Sy, , But pair of 2's and 5's are =2, Then,, , Number of zero's =2, , Il?

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Alternate:Highest power of 5's in 10!, , Number of 5's = 2, Then number of zeroes = 2, Ex,10 Find the number of zeroes at, , the end of the product 100!, Sol. 5, , , , 100, S po 24 (add up all the, Sao quotients), , Number of S's = 20 + 4= 24, then number of zeroes = 24, Bx.11 Find the number of zeroes at, , K the end of the product 1000!., , Be RDS secvses 999 x 1000, 51000, 5 [200 —>, 5[40_ —> \ 249, ss. —, Sls., , , , me highest power of Number, 5's = 200 + 40 + 8 + 1= 249, then number of zeroes at the, end of the product = 249, Ex.12 Find the number of zeroes at, ere end of the product, 1x3x5«7 «. 73 * 1024, Sol. 1*3«x5*7 w 73 * 1024, Number of 5's from 1 to 73, [73, U4 —> \ 16 (add up all the, \2_—> } we, Total number of S's = 14+ 2= L, we know that, 1024 = 2°, , number of 2 = 10, number of pairs (' P10, , then number of, Ex.13 Find the rumah yor at, the end of thi ict,, ens Rots ie 84, Sol. In this, ty; eXpression first, for we the sereis, 1*x2*, sersree 84 FL, , li*12* 13, 1x2 iu, Number of zero (1 to 84), , 5 [84, 5 [16 —> 19, §[3_-—>, , 0, Number of zero (1 to 11), , $l}, sft ys, 0, , Number of zeroes = 19 - 2 = 17, , , , , 5, 5, 5, , , , , , , , , , Sol., , , , Ex.14 Find the number of zeroes at, , the end of the product, , 512 * 513 +1120, , axQx3 Sli x 512 «, 1120- 1%2*3..0., , , , , , 277, , jae, , , , 5, 5, 5, 5, , Number of zeroes = 277 - 126%,, =151 ( <, Ex.15 Find the number of zeroes ats, ‘, , , , the end of the product“{, , 15 x 28 x 38 325A ay, In this type every se@ond, terms has power of 2's. It, means power of 2'$%more than, that of 5 So courit the power, of S's cary,, , =e,, power of 5's(Stotal power of S's, , Sol., , , , Se = (1x Sih b = 5, , 10°= (2x S)heyx ss =5, , 155 = BS =5, , 20° 3“ xis/P asx 55 = 5, D> thee BS =, , 25° “Sloe p)> = 55x 5° = 10, , 30° = x5)= 69x55 =5, , Number of 5's power = 35, ‘tage number of zeroes at the end, of the product = 35, Ex.6 Find the number of zeroes at, the end of the product, J} & 22x 39 2 44 is 28%, , Sol. count the numbe of S's, power of 5's = total power of 5's, 55 = (1x 5)5 = 15x 58 =5, 10%= (2x 5)= 2x51 = 10, 158 = (3x 5)'§= 35x 515 = 15, 20% = (4x 5)? = 4% x 52% = 20, 25% = (5x 5)? = 55x 5% = 50, , Number of 5's power, =5+10+15+20+50, = 100, Then number of zeroes at the, end of product = 100, Ex.17 Find the number of zeroes at, the end of the product, a= 1%, b= 2) 0835, wi. 2= 26%, axbxcxdnn. XZ, Count the number of 5's, power of S's = total power of 5's, , , , Sol., , Rakesh Yadav Readers Publication Pvt. Ltd., , gto= (1x5)? 17* 57 =7, 102 = (2 x 5)!2= 22 5? =12, 1g7= (3x 5)723"x5" = 17, 20% = (4x 5)2= 425% = 29, as = (5x 5)P7=57x 57 = 54, , Number of 5's power, =7+12+17+22+54=11, Then number of zeroes at th, end of product = 112, Ex.18 Find the nymber of zeroes g, the end ofjthe product, 1) 92 Ag? B44... 100:, oe tenis of 5's, , Co, 5 3, , geN6, ais =15, VS, , Sol., , 100' = 100, 5+10+15, it is an a.p.series, , we use a.p. formula, , , , number of term =, , , , , , , , L= last term ofa.p., a= first term of a.p., , , , , , d = common difference, 100-5 |,, number of term = ——> t, = 20, , sum of n™ term of a.p., , = S28 +(n-1)d], , = Fl2x5+(20-1)3], =10[10+19x5], , = 10[105] = 1050, 25% = (5 x SPS = S55 x SS, 50° = (2 x5 x 5 =29 x 5 x 50, TSP = (3X5 x SS = 3% x SS x SS, 100 = (4 5 x 5}! m= 400 x 5100 ¢ 5100, then number of 5's power, =25+50+75 + 100, = 250, then number of total zerc, at the end of product, = 1050 + 250 = 1300, Ex.19Find the number of zeroes, oa the end of the product, 10 x 20 x 30........ 80, 10! x 1 x 101 x 2 10! x 3..B= 108(1% 2% 3... 8), Ie, , but, , Sol.

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ee, , , , , , , , , , , , , , , , , , , , , ise ts from 1 to 8, number of O's=1 Sol. 10'x 1x 10! x 2x 10! 3..,.10!, ae -. Only one pair (2 & 5) x 100, i af then total number of 0's = 10! |1x2%3x,.....100], 54 =1+8=9 from 1 to 100 number of 0's, ee Ex.20 ne o put be zeroes at s]100, e end ol € product *, a 10% 20's 30... 1000 spo} 24 add up all the, TOES at we om quotients), 12, of zeros:, vet" a EXERCISE, -- 100199, f S's 1, Find the number of zeroes at 15. Find the number of zeroes at 26., the end of the product 47! the end of the product, (8 0)9 (c)10 11 21%22%23.......59*60, 2. Find the number of zeroes at (a)14 (b)4 = (c)10 (dy 12, the end of the product 125! 16. Find the number of zeroes at 27., (a) 25 (b) 30 (c) 31 (a) 28 the end of the product, 3. Find the number of zeroes at 35 * 36 « 37 * o.. 89 x 90, the end of the product 378! (a)21_ (b)7 (ce) 14 (a) 20 4, (a) 93 (b) 90 (c) 75 (d) 81 17. Find the number of zeroe, 4. Find the number of zeroes at the end of the product “%,) }, the end of the product 680! 41x 42. Y, {a) 163 (b) 169 (c) 170 (d)165 (a) 26 (b)9 (c) Lk? (d) 25, 5. Find the number of zeroes at 18. Find the number of 2 ‘oes at, the end of the product 1000! the end of the pro a 29,, (a) 200 (b) 249 (c) 248 (d) 250 140!1x5*15%22x 1 fs, : 6. Find the number of zeroes at (a)34 (35, -P- the end of the product 500! 19. Find the“hufnbe, erence (a) 100 (b) 124 (c) 120(d) 125 the end of theoproduct 100-5 , 7: Find the number of zeroes at 25! x 321 3s! i, : +1 the end of the product 1132! {ay 2 (c)22 (d)7, 5 (a) 280 (6) 271 (c) 281(d)272 20. Fi ¢ flumber of zeroes at, = 20 8. Find the number of zeroes at the end of ie Brocuct, pf a.p- the end of the product 1098! : 140F% 358! « 171! ai., (a) 280 (b) 270 (c) 271 (a) 262%, (2) 282 (b) 325 (c) 411 (a) 370, 9. Find the number of zeroes af, 24, Find the number of zeroes at, the end of the product 2346! 4 j) tHe end of the product 1 «2? x, 3? x 4% x 55 6 Ae., 1)x5] (a)580 (b) 583 (c) 575 a (a)225 (b) 250 (c)240 (a)245 32., 10. Find the number of zero iS 22. Find the number of zeroes at, the end of the product, the end of the product, 7 BAe Soom asin Farnam ag, 0 tee se De IS (a) 970 (b) 1124(c) 875 (4) 975, SB 448 98% o J 23. Find the number of zeroes at, Ox Sx 5 (a)2 , ay 5 the end of the product, Sex TERS yy pe, ne Noah ty vo cnae Lie 22! 331 441... 10108 a,, andi 4 the : vd ee ‘che ‘pradiuct (aod. (B10 Meh SL ttt, pernowet eels hot NOT, f 100 ( 3 24, Find the number of zeroes at gg,, e)4 (5 (2a) the end of the product, , f total zerot 3. Find the number of zeroes at, , , , the e roduct, oduct ee of une P, 1300 f)6 (by12 (e)7 (A) S, r of zeroes 914, Find the number of zeroes at, roduct the end of the product, 30 63*5x7x9«11........99*101, | f)24 (5s (2 (do, , , , Q2x 51x 42x 108% 69x 1512x Bx, , 20'¢x 10! 25%, , (a)98 (b)90 (c)94 (d) 100, 28. Find the number of zeroes at the, , end of the product, , 3200+1000+40000+32000+ 15000, , {a)15 (b)13 (c)2 (3, , 36., , Rakesh Yadav Readers Publication Pvt. Ltd., , '20*40x7600%600%300x 1000, he 11, , number of 0's = 20 + 4 = 24, and 10' , here number of zero, = 100, , total number of 0's, , = 24 + 100 = 124, , 124, , then number of zeroes, , , , AAY, Find t! snuthber of zeroes at, the end. of the product, , 3200x 1000x46000x32000~ 15000, , (ants (b)2 (c) 14 (a 16, the number of zeroes at, , chijend of the product, , (b)10 (c)2 (d)3, Find the number of zeroes at, the end of the product 100! +, 200!, , (a)24 (b) 25, (d)N.O.T, , (c) 49, , . Find the number of zeroes at, , the end of the product, ix Q2 x 33x 44 10!°, (a)10 (b)15, (d) N.O.T, Find the number of zeroes at, the end of the product, , (ce) S, , 100! x 200!, , (a)49 (by 24 (c) 73, (d)N.O.T, , Find the number of zeroes at, the end of the product, 5*10*15%x20%25x30x35x40%x45x50, {a8 (b)12 (10 a) 14, , Find the number of zeroes at, the end of the product, 2*4x*6x*8x 10200. 200, (a)49 (b)24 (c)25 (d)SO0, Find the number of zeroes at, the end of the product, 1*3x5x*7, (a)23_ 6 (0 @s, Find The No. zero at the end of, the product of 22? x 5585, , (a)222 (b) 555 (c) 777 (4) 333, Find the number of zeroes at, the end of the product, , 10 + 100 + 1000 +, .--100000000, (a)8_—(b)28 (c)O ad, , Find the number of zeroes at, the end of the product, , TON 103% 103% 104 accsuy, (a)10 (b) 55 (c) 50, , 10", (d) 45,, , Ie

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. Find the number of zeroes at, the end of the product, 2x5 2x29x §4xD8x 59x27 58K QRH, , , , (a) 30 (b) 25, , (c} 55 (a) 50, , 1. (ce) 5. (b) 9. (b), 2. (c) ,6. (b) 10. (a), 3. (2) 7. (c) 11. (b), 4. 0b) 8. (c) 12, (b), , 1. (), , , , 2. {c), , , , , , , , , , 38. Find the number of zeroes at, the end of the product, (3°33- giz. 3'!) (Ds i Q)120 2119), , , , , , —, , {, (g'3 - 3122. B11) (329-3222. 370, (a)1_ (B)2— (c)O_— (3, , 40 Find the number of zeroes at thy, end of the product !, , , , , , , , , (aj1 (bj (ce) 119 (a) 120, 39. Find the number of zeroes at 5x 10x15... 75 |, the end of the product (a)11 (b) 15 (c)10 (dj 18 |, ANSWER KEY |, 13. (a) fe 17. (c) 21. (b) 25. (c) 29. (b) 33. (b) 37. (0)|, 14. (d) By 18. (4) 22. (b) 26. (d) 30. (c) 34, (a) 38. (a), 15. (c) 19. (b) 23. (c) fy 27. (a) 31. (a) 35. (d) 39. (b), 16. (c) 20. (c) 24. (a) 28. (a) 32. (b) 36. (b) 40. (a), SOLUTION, 6. (b) 5 | 500, 5 [100 2*5x*3x5x2*2*11*, 3 [20 i2s x Qx7K2QKSxK7, 0 % 5, In this expression, , ey &%, No. of zeroes = 100 ¥\20) + 4, 2, , , , , , , 7. (e)5 [1132 {, 5 [226 4, , , , , , =25+5+1=31, 3. (a) 5 [378 roes = 226+ 45+9+1, 75 —5 [5 — }ss, s[3—, 0, No. of zeroes = 75 + 15 + 3m 93 ,, *aten 0, 4. (b) 5| 680 “7 No. of zeroes = 219+ 43+8+1, 5[136—- , j =271, 527 Ago’, 9.(b) 5 [2346, SL5 6 jh fy 5 —_, sLi 1 5 —, Am, \ %, 5/18. — {583, No. of eroee 136 + 2745+ 1 3 s 7", ¥ 168 No, of zeroes = 469 + 93 + 18+3, = 583, 5. (b) 5 [1000 10.(a ma, 5 [200 —» An) § | 2700, 5 [40> 5249 5 [108_— \ 673, 5 [8 _* 521 —, s5L_ oe, 9 0, No. of zeroes = 200 + 40 + B+ 1 No. of zeroes = 540 + 108 + 21+ 4, , = 249 = 673, , Rakesh Yadav Readers Publication Pvt. Ltd., , No of 2's = 6, No. of 5's = 3, Pair of 2's and S's = 3, So, No of zeroes = 3, 12.(b) 12x5*x 15*24%13*30%79, 2x2x3x5x3x5*2*, x 2x3x13%2%3x5*S*5*, No. of 2's > 6, No. of 5's > 5, Pair of 2's and 5's = 5S, No. of zeroes = 5, 13.(a) 2x4x6x.......48x50, => 2x1*2x2x2x3 a 2X24x2x!, => 225 (1%2x3x4x.......25), There are many 2's In TH, series we count S's, , 5 [25, , s[—, , $fr}¢, 0, , No. of 5's =5+1=6, Then No. of zeroes = 6, 14, (d) 1*3*5x7x9x1 99x 10, There is no ‘zero’ in tl, expression because there is, even present here., 15.(c) 21*22%23....... 59x60, 123.1 9*20%2 1X2223......59%, Lec, , = 1%23......20, Lo, lit

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5 |60, 2 Fes} 5] 20, \, © 3 faesi4 2 ih, 0 J, , No. of zeroes 1 to60 = 12+2= 14, No. of zeroes 1 to 20 = 4, No. of zeroes 21 to60 = 14-45 10, 16. (c) 35 x 36 x 37 x, 1X2x3%4.....34*35x36......89x90, , , , -1X2%3% 2X3 3x34,, No, of zeroes T to 90 = 18+ 3= 2), , 5, 2 FE x, 0 «5, ] No. of Zeroes 1 to 34, 5 [34, Esp, 0, , No. of Zeroes = 6 + 1 =7, No. of zeroes 35 to 90 = 21-7, , nm, , , , =14, 1T{e) 41 x 42 109 x 110, 1*2%3x4.....40%41%42....., 109x110, , -1*2x3.....40, , No. af zeroes 1 to 110 = 22 +4 = 26, No. of zeroes 1 to 40 =8 + 1=9, , 5 {110 5140, 5[22 > 5[8 Yo, 5[4~+f28 5f1, , 0 0, , No. of zeroes 41 to 110 = 26-9, =17, , 18{d) 140! would have 28 + 5+ 1 =, , 2 5 {140, , 5 [28, ni > = hos, , , , fan, , , , , , 5, 5 [T+ =, Now Remaining part, sri rast Pa, x, , 3x5 x 2x 1, , , , ‘of the Expression, O Would have 3 zeroes, Total No. of zeroes = 34 + 3 = 37, , , , , , hi,, 29-4b) S | 25 5|.32, 5[5 -y 5 S,, sts sft, at Q, ~ sts 1, 7 a, 0, , No. of zeroes In 251=5+1=6, , No of zeroes In 321 = 6+ 1=7, , No. of zero In 4519 + 1 = 10, Total No. of zero = 6 +7 + 10= 23, , , , , , , 20.(c), 5 [1140, 5|358, 2 a 5(71_—, 5 —» (283 5,14 _—» }87, 5 st2—, =, 0, , , , 5 L171, 5 [3¢ —, S [6 > bai, sjlo—, Oo, No. of zeroes in 1140!, = 228+45+9+1=283, , No. of zeroes in 358!, =71+ 14+2=87, , =41, , 21.(b) The Fives will be 1, two's Hence, we, only the Fives, , ), No. of zeroes in 171!, =34+6+1=41 ., Total No. of zeroes= 283+87+1 40000, the, , , , Thus,, Soar 5, 10 = (5.x = 10, 155 = S's 15, 20% 2 = 20, 25% x 5) = 50, 30° (5x 6)2 = 30°, 5 =7 (5x 7) = 35, = (5x8) 40, R45" = (5x9) = 45, , 510+ 15+ 20+50+30+35+40+45,, No. of Fives = 250, Then,, Number of zeroes = 250, , 22.(b) The Five will be less than the, two's Then count the number of, five, 100'* 95®« 90", 10% x 596, (1+6+11.,,....91+96) using sum of, AP,, a=1, d=5, , 96-1, , , , No. of term = +1=20, , S,= 2 lax 1+19%5), , = 10 (2 + 95] = 970, , But, , 100!=(5x5x4)! = 5! x 5!, 752 = (5x5x3)20 = 526 x 526, , Rakesh Yaday Readers Publication Pvt. Ltd., , 6 o*count, , 508) = (5x5x2)5! = 55! x 55!, 257 = (5x5) = §7 x 576, Then no of zeroes, = 1+26+51+76 = 154, Total number of zeroes, = 154+ 970 = 1124, , 23 (c) Count the No. of 5's, , 55 and 10%, No. of 5's = 5! + 40!, Then , e, No. of + 10!, , 24.(a) Cou . of 5's, Then, , 5* x 108 x 1512 x 2016 x 2520, = + 12 + 16 + 40, , 8, , of zero = 80, , 3200, 1000, , 32000, + 15000, 91200, No. or zero = 2, 26.(d) 3200 x 1000 x 40000 x 32000, x 15000, No. of zero's 2+3+4+3+3, =15 ., But 1500 = 3 x 5 x 100, Here 5 is present, When 5 is multiply by even, number, then unit digit '0' is, get., Then,, No. of Total zero = 15 + 1 =16., 27.(a) 20 x 40 x 7600 x 600 x 300, x 1000, No. of zeroes = 1+1+2+2+2+3, =11, 28.(a) 100! + 2001, No. of zeroes In 100! = 20+ 4= 24, No. of zeroes In 200! = 40 +:8 + 1, =49, When you add the two Number, (One with 24 zeroes and the, other with 49 zeroes at It's end), The Total No. of zeroes = 24, 29.(b) 1! 22x 33x qe 19°, Count the Number of 5's, 5° no of fives = 5, 10'° No. of Fives = 10, No. of zeroes = 5 + 10 = 15, , 11